Three, (50 points) Find all positive integers $a, b, m, n$ such that $m$ and $n$ are coprime and $\left(a^{2}+b^{2}\right)^{m}=(a b)^{n}$.


From $a^{2}+b^{2} \geqslant 2 a b>a b$, we know $m1,2^{m}=d^{2(n-m)}$, then $d=2^{r}\left(r \in \mathbf{Z}_{+}\right)$.
Therefore, $a=b=d=2^{r},(2 r+1) m=2 r n$.
But $(2 r+1,2 r)=1,(m, n)=1$, thus, $m=2 r, n=2 r+1$.
Hence, $(a, b, m, n)=\left(2^{r}, 2^{r}, 2 r, 2 r+1\right)$, where $r$ is a positive integer.