As you know, $x = 2$, $y = 4$ is the only solution in integers. Well, also $x = 4$, $y = 2$ if you want to be a smart aleck about it.
But what if you expand the search to include $x$ and $y$ that are rational, or irrational, or imaginary, or complex? Are any other solutions to be found, keeping only the restriction $x \neq y$?
What I have done to try to solve it: just some queries on Wolfram Alpha.
Here is a standard parameterization:
Let $y = rx$. Then $x^y = y^x$ becomes $x^{rx} = (rx)^x$.
Taking $x$-th roots, $x^r = rx$ or $x^{r-1} = r$ or, finally, $x = r^{1/(r-1)}$.
From this, $y = rx =r^{1+1/(r-1)} =r^{r/(r-1)} $.
Put in any value of $r$, you will get an $x$ and $y$ satisfying $x^y = y^x$.
Note: My answer here (Are there real solutions to $x^y = y^x = 3$ where $y \neq x$?) shows that the smallest positive real value of $z$ such that there are distinct $x$ and $y$ such that $x^y = y^x$ is $z = e^e \approx 15.15426224 $.
This is why $2^4 = 4^2 = 16$ works.