Finding primes so that xp+yp=zpx^p+y^p=z^p is unsolvable in the p-adic units

On my number theory exam yesterday, we had the following interesting problem related to Fermat’s last theorem:

Suppose $$p>2p>2$$ is a prime. Show that $$xp+yp=zpx^p+y^p=z^p$$ has a solution in $$Z×p\mathbb{Z}_p^{\times}$$ if and only if there exists an integer $$aa$$ such that $$p∤p\not\mid a(a+1)$$ and $$(a+1)^p=a^p+1\pmod{p^2}.(a+1)^p=a^p+1\pmod{p^2}.$$
(I wil not show this here, it’s a good exercise in Hensel’s lemma)

Now I’m interested for which primes this last property holds true.
This is what I found so far:

Proposition: for every prime $$p\equiv 1\pmod{3}p\equiv 1\pmod{3}$$ there exists an $$a\in\mathbb{Z}a\in\mathbb{Z}$$ such that $$p\not\mid a(a+1)p\not\mid a(a+1)$$ and $$p^2\mid(a+1)^p-a^p-1p^2\mid(a+1)^p-a^p-1$$.
Sketch of proof: We will often use the fact that if $$p\mid a-bp\mid a-b$$ then $$p^2\mid a^p-b^pp^2\mid a^p-b^p$$ for $$a,b,p\in\mathbb{Z}a,b,p\in\mathbb{Z}$$. There exists some $$n\in\mathbb{Z}n\in\mathbb{Z}$$ such that $$p \mid n^2+n+1p \mid n^2+n+1$$. We will show that we can take $$a=na=n$$ (it is clear that $$p\not \mid n(n+1)p\not \mid n(n+1)$$). If $$n\equiv 1\pmod{p}n\equiv 1\pmod{p}$$, we are done because $$(n+1)^p\equiv 2\equiv n^p+1\pmod{p^2}(n+1)^p\equiv 2\equiv n^p+1\pmod{p^2}$$. So we may assume that the order of $$nn$$ mod $$pp$$ is 3.

We see that $$n^{3p}\equiv 1\pmod{p^2}n^{3p}\equiv 1\pmod{p^2}$$, hence $$p^2\mid \frac{n^{3p}-1}{n^p-1}=n^{2p}+n^p+1p^2\mid \frac{n^{3p}-1}{n^p-1}=n^{2p}+n^p+1$$ and therefore $$(n+1)^p\equiv -n^{2p}\equiv n^p+1\pmod{p^2}(n+1)^p\equiv -n^{2p}\equiv n^p+1\pmod{p^2}$$. $$\blacksquare\blacksquare$$

What can we say about primes $$p\equiv 2\pmod{3}p\equiv 2\pmod{3}$$? Say a prime is bad if it satisfies the above property. It is easy to see that for $$pp$$ to be good we only have to check that for all $$1\le a \le (p-1)/21\le a \le (p-1)/2$$ we have $$p^2\not\mid (a+1)^p-a^p-1p^2\not\mid (a+1)^p-a^p-1$$.
Numerical data suggests that infinitely many primes $$p\equiv 2\pmod{3}p\equiv 2\pmod{3}$$ are good and infinitely many are bad. I’d like to find a proof of this.
My numerical data furthermore suggests that there is no easy divisibility criterion to judge badness for primes, and that that there are a lot more good primes than bad ones, in terms of asymptotics.

Edit: Here are some observations. I’ve listed the first bad primes, together with the set of$$a\in\mathbb{Z}a\in\mathbb{Z}$$ such that $$1\le a \le (p-1)/21\le a \le (p-1)/2$$ and $$(a+1)^p\equiv a^p+1\pmod{p^2}(a+1)^p\equiv a^p+1\pmod{p^2}$$:

$$59\qquad [3, 4, 11, 14, 15, 20] \\ 83\qquad [8, 30, 36] \\ 179\qquad [2, 59, 88] \\ 227\qquad [36, 82, 92] \\ 419\qquad[80, 110, 150] \\ 443\qquad [108, 125, 201] \\ 701\qquad [23, 61, 132, 146, 153, 252] \\ 857\qquad [6, 143, 244] \\ 887\qquad [20, 132, 168] \\ 911\qquad [14, 64, 242] \\ 929 \qquad[234, 253, 266] \\ 971 \qquad[9, 97, 108] \\ 977 \qquad[102, 182, 331] \\ 1091\qquad [64, 234, 358] \\ 1109 \qquad[176, 400, 523] \\ 1193\qquad [224, 227, 473] \\ 1217\qquad [186, 228, 410] \\ 1223\qquad [304, 349, 409] \\ 1259\qquad [19, 62, 264] \\ 1283\qquad [19, 134, 449] \\ 1289\qquad [412, 535, 593] \\ 1439\qquad [93, 199, 294] \\ 1487\qquad [167, 416, 649] \\ 1493 \qquad[141, 179, 367] \\ 1613 \qquad[227, 473, 739] \\ 1637 \qquad[76, 279, 574] \\ 1811 \qquad[39, 123, 498, 628, 691, 743] \\ 1847\qquad [172, 362, 698] \\ 1901\qquad [3, 475, 634] \\ 1997\qquad [125, 671, 840] \\ 2003 \qquad[31, 312, 840] \\ 2087 \qquad[31, 202, 586] \\ 2243 \qquad[252, 593, 1077] \\ 2423\qquad [353, 704, 857] \\ 2477\qquad [17, 688, 1020] \\ 2579 \qquad[294, 576, 693] \\ 2591 \qquad[322, 345, 368] \\ 2729\qquad [1024, 1049, 1089] \\ 2777\qquad [488, 1078, 1269] \\ 2969\qquad [341, 789, 1158] \\ 3089\qquad [677, 1015, 1053]\\ 3137\qquad [1227, 1308, 1427]\\ 3191\qquad [776, 916, 1383]\\ 3203\qquad [1214, 1305, 1587]\\ 3251\qquad [164, 337, 1260]59\qquad [3, 4, 11, 14, 15, 20] \\ 83\qquad [8, 30, 36] \\ 179\qquad [2, 59, 88] \\ 227\qquad [36, 82, 92] \\ 419\qquad[80, 110, 150] \\ 443\qquad [108, 125, 201] \\ 701\qquad [23, 61, 132, 146, 153, 252] \\ 857\qquad [6, 143, 244] \\ 887\qquad [20, 132, 168] \\ 911\qquad [14, 64, 242] \\ 929 \qquad[234, 253, 266] \\ 971 \qquad[9, 97, 108] \\ 977 \qquad[102, 182, 331] \\ 1091\qquad [64, 234, 358] \\ 1109 \qquad[176, 400, 523] \\ 1193\qquad [224, 227, 473] \\ 1217\qquad [186, 228, 410] \\ 1223\qquad [304, 349, 409] \\ 1259\qquad [19, 62, 264] \\ 1283\qquad [19, 134, 449] \\ 1289\qquad [412, 535, 593] \\ 1439\qquad [93, 199, 294] \\ 1487\qquad [167, 416, 649] \\ 1493 \qquad[141, 179, 367] \\ 1613 \qquad[227, 473, 739] \\ 1637 \qquad[76, 279, 574] \\ 1811 \qquad[39, 123, 498, 628, 691, 743] \\ 1847\qquad [172, 362, 698] \\ 1901\qquad [3, 475, 634] \\ 1997\qquad [125, 671, 840] \\ 2003 \qquad[31, 312, 840] \\ 2087 \qquad[31, 202, 586] \\ 2243 \qquad[252, 593, 1077] \\ 2423\qquad [353, 704, 857] \\ 2477\qquad [17, 688, 1020] \\ 2579 \qquad[294, 576, 693] \\ 2591 \qquad[322, 345, 368] \\ 2729\qquad [1024, 1049, 1089] \\ 2777\qquad [488, 1078, 1269] \\ 2969\qquad [341, 789, 1158] \\ 3089\qquad [677, 1015, 1053]\\ 3137\qquad [1227, 1308, 1427]\\ 3191\qquad [776, 916, 1383]\\ 3203\qquad [1214, 1305, 1587]\\ 3251\qquad [164, 337, 1260]$$

I also found this sequence on OEIS, along with more numerical data, where it is claimed (but not proved) that the density of bad primes is roughly $$\frac 16\frac 16$$.

My numerical data also suggests that the number of $$a\in\mathbb{Z}a\in\mathbb{Z}$$ satisfying $$1\le a \le (p-1)/21\le a \le (p-1)/2$$ and $$p^2\mid (a+1)^p-a^p-1p^2\mid (a+1)^p-a^p-1$$ is always either $$33$$ or $$66$$, but I think that’s rather a bold statement. (edit: indeed, 20123 is a counterexample with $$99$$ solutions, found by Michael Stocker. The weaker statement that the number of such solutions is divisible by $$33$$ is true, however: see here.)

Edit: Someone showed me these slides, where the above criterion for solvability of $$x^p+y^p=z^px^p+y^p=z^p$$ in the p-adic units is stated. (There is a remark that it can be proven with eisenstein reciprocity that for all non-Wieferich primes the first case of FLT holds true – maybe there is some way to use this for some (partial) result on our problem? I’m really curious.)