Solve the equation in positive integers $$k^m+m^n=kmn$$
My work so far:
1) Let $k=m$. Then $$k^k+k^n=k^2n$$ If $k=1$ then $n=2$; if $k=2$ then $n=2$ or $n=3$
Let $k\ge 2$. We see that $n\ge2$.
$$k^{k-2}+k^{n-2}=n$$
2) Let $k=n$. Then $$k^m+m^k=k^2m$$. I find $m=2$, $k=2$ or $k=4$
3) Let $m=n$. Then $$k^m+m^k=km^2$$. Similarly.
Suppose first that $k$ is the largest among $k$, $m$, and $n$. Then, $$k^m<k^m+m^n=kmn\leq k^3$$ implies that $k>1$ and $m<3$. If $m=1$, then we have $k+1=kn$, which is not possible (recalling that $k>1$). If $m=2$, then $k^2+2^n=2kn$ implies that $$0\leq (k-n)^2=n^2-2^n\,,$$ yielding $n\in \{2,3,4\}$. Hence, in this case where $k$ is the largest, the solutions in the form $(k,m,n)$ are $(2,2,2)$, $(4,2,3)$, and $(4,2,4)$.
Now, suppose that $m$ is the largest of the three variables. Then, $$m^n<k^m+m^n=kmn\leq m^3 $$ implies that $m>1$ and $n<3$. We start with the subcase $n=1$, which leads to $$k^m+m=km\text{ or }k^m = (k-1)m\,.$$ From the equation above, $k-1$ divides $k^m$, which only happens when $k=2$. Nonetheless, this leads to a contradiction $2^m=m$. Thus, $n=2$ is the only hope here. Plugging $n=2$ into the original equation, we get $$k^m+m^2=2km\text{ or }0\leq (m-k)^2=k^2-k^m\,.$$ This proves that $k=1$ or $m=2$. If $k=1$, then $m=1<2=n$, which is absurd. Thus, $m=2$ and we obtain $(k,m,n)=(2,2,2)$.
Finally, we assume that $n$ is the largest. We note that $$m^n<k^m+m^n=kmn\leq n^3\,.$$ The inequality above implies that $m<4$. There are a few easy-to-check scenarios listed below.
In summary, there are five solutions $(k,m,n)\in \left(\mathbb{Z}_{>0}\right)^3$ to $k^m+m^n=kmn$. All solutions in the form $(k,m,n)$ are $(1,1,2)$, $(2,2,2)$, $(2,2,3)$, $(4,2,3)$, and $(4,2,4)$.