I wrote about this in another post.
The other solution is 4.
For $x^a = a^x; a > 0$ in general... Well, notice the shapes of the graphs for $x \ge 0$ are such that they intersect twice if $a \ne e$ and intersect once if $a = 1$ (at $x = e$).
In my other post I went into great detail about taking the first and second derivatives to show both $a^x$ and $x^a$ are concave and that they can only have at most two intersections and as $a^0 = 1 > 0^a = 0$ there must have at least one. But I'll leave that as an excercise to the reader this time.
But obviously $x = a \implies a^a = a^a$ is a solution. The other ... hmm... I forget and I'm too lazy to figure it out a second time. But the solutions are "paired" (that is if $x = b$ solves $a^x = x^a$ then $x = a$ solves $b^x = x^b$) and one solution is less than e and the other is greater than e. I came up with a formula relating the solutions.
2 and 4 solves both $x^2 = 2^x$ and $x^4 = 4^x$ were the slickest solutions but every other positive value for $a$ and a $a$ and $a'$ solution.
If $a$ is even or an even rational then there is an $x = -1/a$ solution. But if a is odd then there is no negative solution as for $x < 0$ then $a^x > 0$ but $x^a < 0$. If a is irrational then $x^a$ is undefined for $x < 0$. These are the only solutions as $a^x$ is increasing but $x^a$ is decreasing when $x < 0$ and $a$ is even.