We start with a lemma.
Lemma 1. Let $n$ be a positive integer and $p, q$ some positive rational numbers. If $n^p = q$, then $q$ is itself an integer.
Proof. Suppose $p = \frac{a}{b}$ and $q = \frac{c}{d}$ where $a, b, c, d \in \mathbb{N}$. We have
$$n^p = q \Rightarrow n^{\frac{a}{b}} = \frac{c}{d} \Rightarrow n^a = \left(\frac{c}{d}\right)^b = \frac{c^b}{d^b} \Rightarrow d^b \mid c^b \Rightarrow d \mid c \Rightarrow q \in \mathbb{N}$$
□
Now for the main problem, note that if $n = 1$, then $m = n = 1$ and this is a solution for the equation. So we may assume that $n > 1$. Let $r = \log_n m$, ($m = n^r$). We have
$$n^{nn} = m^m = (n^r)^{nr} = n^{rnr}$$
Since $n > 1$, we must have
$$n^n = rn^r \Rightarrow r = n^{n-r}$$
On the other hand, since $n^{n^n} = m^m$, we get $n^n = m \log_n m = r$ or $r = \frac{n^n}{m} \in \mathbb{Q}$. According to the lemma $n - r$ and $r$ playing the role of $p$ and $q$, respectively, we get $r$ is an integer. Now if $r < n$, then $n^{n-r} \ge n^1 > r = n^{n-r}$, which is impossible. And if $r > n$, then $n^{n-r} < 1 \le r = n^{n-r}$ which is again impossible.
So we must have $n = r$. Hence, $n = r = n^{n-r} = 1$. This contradicts with the assumption $n > 1$ and consequently, the only solution is $m = n = 1$.