How to solve these two simultaneous “divisibilities” : n+1∣m2+1n+1\mid m^2+1 and m+1∣n2+1m+1\mid n^2+1

Some further results along the lines of thought of @individ:

Suppose $p$ and $s$ are solutions to the Pell’s equation:
$$-d\cdot p^2+s^2=1$$
Then,
$$\begin{align} m &= a\cdot p^2+b\cdot pq +c\cdot q^2\\ n &= a\cdot p^2-b\cdot pq +c\cdot q^2 \end{align}$$
are solutions if $(a,b,c,d)$ are: (these are the only sets that I found using the computer)
$$\begin{align} (10,4,-2,-15)\\ (39,12,-3,-65)\\ \end{align}$$
Sadly, the solutions are negative.

Here are some examples:
$$\begin{align} (m,n) &= (-6,-38) &(a,b,c,d,p,q)&=(10,4,-2,-15,1,4)\\ (m,n) &= (-290,-2274) &(a,b,c,d,p,q)&=(10,4,-2,-15,8,31)\\ (m,n) &= (-15171,-64707) &(a,b,c,d,p,q)&=(39,12,-3,-65,16,129)\\ (m,n) &= (-1009692291,-4306907523) &(a,b,c,d,p,q)&=(39,12,-3,-65,4128,33281)\\ (m,n) &= (-67207138138563,-286676378361411) &(a,b,c,d,p,q)&=(39,12,-3,-65,1065008,8586369)\\ \end{align}$$
P.S. I am also very curious how @individ thought of this parametrization.