Checking a Proof of a Theorem

Theorem 1.2 of Bennett and Skinner (Canad. J. Math., 2004) asserts that the Diophantine equation $x^{p} - 4y^{p} = z^{2}$ is unsolvable for every prime $p \geq 7.$
The following is a possible proof (from an arXiv author) that Fermat’s Last Theorem is a consequence of this theorem, i.e., proof that if there exist integers $x, y, z > 0$ such that $(x, y) = 1$ and $x^{p} + y^{p} = z^{p},$ then there exist integers $a, b, c > 0$ such that $(a, b)=1$ and $a^{p} - 4b^{p} = c^{2}$.

Take a prime
$p \geq 7.$
We will prove Fermat’s Last Theorem in the form:
Take integers $x, y, z > 0$.
If
$(x, y) = 1,$
then
$x^{p} + y^{p} \neq z^{p}$.

We argue by contradiction.
By the equation
$x^{p} + y^{p} = z^{p}$
there is a rational
$0 < r < 1$
such that
$$x^{p} = rz^{p}\ \ \mbox{and}\ \ y^{p} = (1-r)z^{p},$$
so that
$$r^{2} - r + \dfrac{(xy)^{p}}{z^{2p}} = 0,$$
and hence
$$r = \dfrac{1 + \sqrt{1 - 4(xy)^{p}/z^{2p}}}{2}\ \ \mbox{or}\ \ \dfrac{1 - \sqrt{1 - 4(xy)^{p}/z^{2p}}}{2}.$$
Therefore,
the difference
$1 - 4(xy)^{p}/z^{2p} \geq 0$
is to be a perfect square.
But since
$$1 - \dfrac{4(xy)^{p}}{z^{2p}} = \dfrac{z^{2p} - 4(xy)^{p}}{z^{2p}},$$
since
$z^{2p}$
is a perfect square,
and since if $z^{2p} = 4(xy)^{p}$
then from the equation
$x^{p} + y^{p} = z^{p}$
we have
$x = y$
that leads to a contradiction,
so there is an integer
$c > 0$
such that
$$z^{2p} - 4(xy)^{p} = c^{2}.$$
On choosing
$$a := z^{2}\ \ \mbox{and}\ \ b := xy$$
we have
$$a, b > 0\ \ \mbox{and}\ \ a^{p} - 4b^{p} = c^{2}.$$
Moreover,
because
$(x, y) = 1$
and
$x^{p} + y^{p} = z^{p},$
we have
$(x, y) = (y, z) = (x, z) = 1,$
whence
$$(a, b) = 1.$$
But the existence of such
$a, b, c$
contradicts Theorem 1.2 of Bennett and Skinner [1].

Answer

Attribution
Source : Link , Question Author : Megadeth , Answer Author : TonyK