[sp]If a $p$th power is also a $q$th power (with $p$ and $q$ relatively prime), then it must be a $pq$th power. So if $(x^2+y^2)^p=(xy)^q$, then there must be some integer $z$ with $x^2+y^2 = z^q$ and $xy = z^p.$ Solving the second of those equations for $y$ and substituting that into the first equation, you get $$x^2 + \frac{z^{2p}}{x^p} = z^q.$$ Therefore $x^4 - z^qx^2 + z^{2p} = 0,$ a quadratic in $x^2$ with solutions $x^2 = \frac12\bigl(z^q \pm\sqrt{z^{2q} - 4z^{2p}}\bigr).$ For the expression in the square root to be non-negative, we must have $q>p$, so that $x^2 = \frac12\bigl(z^q \pm z^{2p}\sqrt{z^{2(q-p)} - 4}\bigr).$ But the only way for two squares to differ by $4$ is if they are $4$ and $0$. Therefore $z^{2(q-p)} = 4$, from which $q = p+1$ and $z=2$. The solution is then $q= p+1$, $x=y=2^{p/2}$, which implies that $p$ must be even. For odd $p$ there is no solution.[/sp]