Proof: Let $(m, n)=d$, then $m=m_{1} d, n=n_{1} d$, where $\left(m_{1}, n_{1}\right)=1$. Thus, the given condition becomes $m_{1} n_{1} \mid\left(m_{1}^{2}+n_{1}^{2}\right)$, hence we also have $m_{1} \mid\left(m_{1}^{2}+n_{1}^{2}\right)$, which implies $m_{1} \mid n_{1}^{2}$. But $\left(m_{1}, n_{1}\right)=1$, so $\left(m_{1}, n_{1}^{2}\right)=1$. Combining $m_{1} \mid n_{1}^{2}$, we must have $m_{1}=1$. Similarly, $n_{1}=1$, therefore $m=n$.