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 arXiv) 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].
