, 2022

Let $\ell$ be a positive integer. By substituting $(m, n) = (f(\ell), \ell)$ and $(\ell, f(\ell))$ into the original equation and comparing them, we obtain
$$f^{f(\ell)+1}(\ell) + \ell f(\ell) = f(\ell)f(f(\ell)) = f^{f(f(\ell))}(\ell) + \ell f(\ell),$$
or $f^{f(\ell)+1}(\ell) = f^{f(f(\ell))}(\ell)$.
Letting $m = n$ in the original equation yield $f(n)^2 = n^2 + f^{f(n)}(n) > n^2$, or $f(n) > n$. Hence, $f^{k+1}(n) = f(f^k(n)) > f^k(n)$ for any positive integer $k$, which leads to
$$f(n) < f^2(n) < f^3(n) < \dots$$
In particular, if $f^s(n) = f^t(n)$ for some positive integers $s, t$, then $s = t$. Combined with $f^{f(\ell)+1}(\ell) = f^{f(f(\ell))}(\ell)$, we obtain $f(f(\ell)) = f(\ell) + 1$.

In particular, by letting $k = f(n)$ in the above statement, $f^{f(n)}(n) = f(n) + f(n) - 1 = 2f(n) - 1$. We also have $f(n)^2 = n^2 + f^{f(n)}(n)$ (see the second paragraph), and we deduce $(f(n) - 1)^2 = n^2$. Since $f(n) - 1 \ge 0$, it implies $f(n) - 1 = n$, or $f(n) = n + 1$.
We conclude the proof by checking $f(n) = n + 1$ satisfies the original equation. The left side of the equation is $f^{f(n)}(m) + mn = m + f(n) + mn = mn + m + n + 1$; the right is $f(m)f(n) = (m+1)(n+1) = mn + m + n + 1$. Therefore, the answer is $f(n) = n + 1$.