Let $p$ be a prime with $p\equiv 3\pmod 8$ , then all rational solutions to $px^4-4y^4=z^2$ satisfy $xyz=0$.
Apparently, the argument to prove this is that $-1$ is not a quadratic residue of $p$.
I know that $-1$ is not a quadratic residue of $p$ because of the law of quadratic reciprocity and $p\equiv 3mod4$. However, I fail to see how this proves that all solutions satisfy xyz=0. If I take the equation modulo $p$, then I'm stuck with $-4y^4\equiv z^2 mod p$. What am I missing?