$12 \cdot 153$ For a given positive integer $m$, find all positive integer triples $(n, x, y)$, where $m, n$ are coprime, and satisfy
$$
\left(x^{2}+y^{2}\right)^{m}=(x y)^{n} .
$$
(53rd American Putnam Mathematical Competition, 1992)
[Solution] Let $(n, x, y)$ be a set of positive integer solutions to the equation.
By the arithmetic-geometric mean inequality, we have
$$
(x y)^{n}=\left(x^{2}+y^{2}\right)^{m} \geqslant(2 x y)^{m} .
$$

Therefore, $n>m$.
Let $p$ be a prime number, and $p^{a}\left\|x, p^{b}\right\| y$. Then
$$
p^{(a+b)^{n}} \|(x y)^{n} \text {. }
$$

If $a < b$, then $p^{(a+b)^{n}} \|(x y)^{n}$, but $p^{(2b)^{m}} \|(2xy)^{m}$, which implies $(a+b)^{n} \geq (2b)^{m}$. Since $n > m$, this leads to a contradiction.
Similarly, if $a > b$, we also derive a contradiction.
Thus, for all prime numbers $p$, we have $a=b$.
Therefore, we can conclude that $x=y$.
Hence, the given equation simplifies to $\left(2 x^{2}\right)^{m}=x^{2 n}$
which is $x^{2(n-m)}=2^{m}$.
This indicates that $x$ is an integer power of 2.
Assume $x=2^{a}$. Then $2^{2 a(n-m)}=2^{m}$
which gives $2 a(n-m)=m$, or $2 a n=m(2 a+1)$.
Since $(m, n)=(2 a, 2 a+1)=1$,
it must be that $m=2 a, n=2 a+1$.
Thus, when $m$ is odd, the equation has no solution.
When $m$ is even, the equation has solutions
$$
(n, x, y)=\left(m+1, x^{\frac{m}{2}}, y^{\frac{m}{2}}\right) .
$$