Prove that using the fact that perfect squares $\equiv 0,1(\bmod 3)$, if $x$ and $y$ are not multiples of 3, then

this leads to
$$\begin{array}{c} x^{2}+y^{2} \equiv 2(\bmod 3) \\ z^{2} \equiv 2(\bmod 3) \end{array}$$

which is a contradiction. Hence, one of $x$ or $y$ must be a multiple of 3.
If $x$, $y$, and $z$ are not multiples of 5, then
$$x^{2}+y^{2} \equiv 0,2 \text { or } 3(\bmod 5),$$

while
$$z^{2} \equiv 1 \text { or } 4(\bmod 5),$$

which is a contradiction. Hence, one of $x$, $y$, or $z$ must be a multiple of 5.
If $x$, $y$, and $z$ are all even, then divide both sides of (1) by 4 until one of $x$ or $y$ is odd. Suppose $x$ is odd, then $y$ must be even (otherwise $z^{2}=x^{2}+y^{2} \equiv 1+1 \equiv 2(\bmod 4)$, which is a contradiction), in this case $z$ is odd. Taking both sides of (1) modulo 8, we find that $y^{2} \equiv 0(\bmod 8)$, hence $4 \mid y$.

In summary, the proposition holds.