Autumn tournament

There are no solutions in positive integers. Without loss of generality, assume that $m\ge n$. If $n\ge 3$, then $${\left(1+\frac{1}{m}\right)}^m=\sum _{k=0}^m\left(\genfrac{}{}{0ex}{}{m}{k}\right)\frac{1}{m^k}<\sum _{k=0}^m\frac{1}{k!}<\sum _{k=0}^{\infty }\frac{1}{k!}=e<n⟹n^{\frac{1}{m}}>1+\frac{1}{m}$$ 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. $\square$