Note that $p > 3$ and $p \mid x^3 - y^3 = (x-y)(x^2 + xy + y^2)$. Since $x, y < p$, $p \nmid x-y$. Therefore $p \mid x^2 + xy + y^2$. Similarly $p \mid x^2 + xz + z^2$, $p \mid y^2 + yz + z^2$.

Thus $p \mid (x^2 + xy + y^2) - (y^2 + yz + z^2) = (x-z)(x+y+z)$ and so $p \mid x+y+z$.

Since $x, y, z < p$ and $p$ is a prime, $x+y+z = p$ or $2p$.

Since the parity of $x+y+z$ and $x^2+y^2+z^2$ are the same, it suffices to prove that $p \mid x^2+y^2+z^2$.

Now $p \mid x^2 + xy + y^2 = x(x+y+z) + y^2 - xz$ and so $p \mid y^2 - xz$. Therefore $p \mid (x^2 + xy + y^2) + (y^2 - xz) = x^2 + y^2 + z^2$ and we are done.