Let $m$ be a given positive integer. Find all positive integer solutions of the equation $\left(x^{2}+y^{2}\right)^{m}=(x y)^{z}$.
According to the arithmetic and geometric mean inequality:

$$
\frac{x^{2}+y^{2}}{2} \geq x y
$$

thus $(x y)^{z}=\left(x^{2}+y^{2}\right)^{m} \geq 2^{m}(x y)^{m},(x y)^{\frac{z-m}{m}} \geq 2$; therefore, necessarily $z>m$. Let the greatest common divisor of $x$ and $y$ be denoted by $d$; $x=a d$ and $y=b d$, where $a$ and $b$ are coprime. Substituting this into the original equation, we get:

$$
\left(a^{2} d^{2}+b^{2} d^{2}\right)^{m}=\left(a b d^{2}\right)^{z}
$$

rearranging

$$
\left(a^{2}+b^{2}\right)^{m}=(a b)^{z} d^{2(z-m)}
$$

Suppose $a \neq 1$, that is, there exists a prime $p$ that divides $a$. Then $p$ is a divisor of the right-hand side of the above equation, and thus of the left-hand side as well. This can only be true if $p$ divides $\left(a^{2}+b^{2}\right)$, and thus $p$ divides $b$. Therefore, $a$ and $b$ have a common prime divisor, which is a contradiction. Hence, $a=1$, and similarly, it can be shown that $b=1$.

Then (1) becomes:

$$
2^{m}=d^{2(z-m)}
$$

Therefore, $d$ must be a power of 2; let $d=2^{k}$. Substituting this in: $m=2(z-m) k$, that is, $m=\frac{2 k}{1+2 k} z .2 k$ and $1+2 k$ are coprime, $m$ is an integer, so $1+2 k$ divides $z$; let $z=(1+2 k) n$. Then $m=2 k n$; thus the problem has a solution only for even $m$, and every solution is of the form $x=y=2^{k}, z=(1+2 k) n$, where $m=2 k n$.