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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.

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    Rational solutions may be found through parametrization of the solution. – 2017-02-07
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    There are a huge number of solutions in the real. $f(x)=x^{1/x}$ has a maximum when $x=e$ and as $x\to 1$, $f(x)\to 1$ and as $x\to \infty$, $f(x)\to 1$. This means for any $a$ with $1e$ so that $x^{1/x}=a=y^{1/y}$, or $x^y=y^x$. – 2017-02-07
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    See the answers of [$x^y = y^x$ for integers $x$ and $y$](http://math.stackexchange.com/questions/9505/xy-yx-for-integers-x-and-y) – 2017-02-07

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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.

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    Alright, I just tried it with $r = 3$ and $r = i$, it checks out. That's enough for me tonight. See you tomorrow. – 2017-02-07
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    And it should be noted, the famous $x = 2$, $y = 4$, comes from $r = 2$. – 2017-02-07