The Diophantine equation $x^p - 4y^p = z^2$ with $(x, y) = 1$ and $x, y, z > p.$

Question

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

Answer 1

This equation is a special case of the generalized Fermat equation $$ Ax^p+By^q=Cz^r $$ for $A=C=1$, $B=-4$ and $p=q$, $r=2$. We have $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}<1$ in our case $(p,q,r)=(p,p,2)$ for $p>3$, so that we are in the hyperbolic case. Hence for $p>3$ there are at most finitely many coprime solutions $(x,y,z)$ by the Darmon-Granville theorem. And probably there are very few such solutions - the abc-conjecture implies that there are at most $2$ solutions once $n>n_0$, independent of $A,B,C$. The equation is not hyperbolic for $p=3$, but the argument with the abc-conjecture still applies. There is a large literature on the generalized Fermat equation, which will be helpful to study this case (the experts might know more).

Edit: A reference was given afterwards here, and Gerry Myerson found $78^3-4\cdot 29^3=614^2$ and $93^3-4\cdot 53^3=457^2$for $p=3$.