TITLE: The Diophantine equation $x^p - 4y^p = z^2$
QUESTION [7 upvotes]: If $p \geq 5$ is a prime, are there any integers $x, y, z > p$ such that
$(x, y) = 1$ 
and
$$x^{p} - 4y^{p} = z^{2}$$

REPLY [10 votes]: See Theorem 1.2 of the paper by Bennett and Skinner, which settles the problem for $p\ge 7$ (take there $C=1$ and $\alpha_0=2$).   Note that the Bennett-Skinner results are more general. (Earlier work of Darmon and Granville (using Faltings's theorem) showed that there are only finitely many solutions; again for more general such equations.)
Finally GH from MO has kindly pointed out an earlier paper of Darmon that handles this particular equation (assuming Shimura-Taniyama) for $p=11$ or $p\ge 17$.