There are no solutions in positive integers. Without loss of generality, assume that $m \ge n$. If $n \ge 3$, then
$$\left(1 + \frac{1}{m}\right)^m = \sum_{k=0}^{m} \binom{m}{k} \frac{1}{m^k} < \sum_{k=0}^{m} \frac{1}{k!} < \sum_{k=0}^{\infty} \frac{1}{k!} = e < n \implies n^{\frac{1}{m}} > 1 + \frac{1}{m}$$
Similarly, $m^{\frac{1}{n}} > 1 + \frac{1}{n}$, whence
$$m^{\frac{1}{n}} + n^{\frac{1}{m}} > 2 + \frac{1}{m} + \frac{1}{n} \ge 2 + \frac{2}{mn} > 2 + \frac{2}{mn(m+n)^{\frac{1}{m}+\frac{1}{n}}}.$$
The cases $n = 1, 2$ can be solved directly. $\square$