If $p \geq 5$ is a prime, are there any integers $x, y, z > p$ such that
$(x, y) = 1$ 
and
$$x^{p} - 4y^{p} = z^{2}$$



The Diophantine equation $x^p - 4y^p = z^2$
See Theorem 1.2 of the paper by [Bennett and Skinner](https://www.math.ubc.ca/~bennett/BS.pdf), 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](http://www.math.mcgill.ca/darmon/pub/Articles/Research/12.Granville/paper.pdf) (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](http://www.math.mcgill.ca/darmon/pub/Articles/Research/08.Fermat-nn23/paper.pdf) that handles this particular equation (assuming Shimura-Taniyama) for $p=11$ or $p\ge 17$. 


