Let $\frac{mn}{m+n} = a$. Then we get that $(m-a)(n-a)=a^2$. This means that $n \ge a+1$. It also means that $m=\frac{an}{n-a}$. We want to prove that $m+n \le n^2$, which we can substitue our value of $m+n$ and after simplifying get that we want to prove that $m \le an$. Now plugging in our value for m in terms of a and n, we get that this is satisfied if $n \ge a+1$, which we know to be true.

We are given that $m + n \mid mn$. Thus, $m + n \mid mn - n(m+n) \Rightarrow n+m \mid -n^2$. Which means that $n+m \mid n^2 $. This directly implies that $m + n \leq n^2$. $\blacksquare$