Commutativity of powers in $\Bbb Q$. The equation $x^y=y^x$ for $x< y$ has just one solution in $\Bbb Z$ ($x=2,y=4$), and infinitely many in $\Bbb R$, but what happens with $\Bbb Q$? And with $\overline{\Bbb Q}$?
I have visited Gelfond-Schneider theorem but it doesn't seem help here.
My try:
Let $(a,b)=(p,q)=1$ where $(x,y)$ denotes the gcd of $x$ and $y$. If
$$\left(\frac ab\right)^{p/q}=\left(\frac pq\right)^{a/b}$$
then
$$\left(\frac ab\right)^{bp}=\left(\frac pq\right)^{aq}$$
Since both fractions are in their lowest terms,
$$a^{bp}=p^{aq}$$
$$b^{bp}=q^{aq}$$
Or
$$(a^b)^p=(p^q)^a$$
 A: Let's start with $\Bbb Q$.
So you've taken $x=a/b,\,y=p/q$. Any prime factor of $a$ ($b$) divides $p$ ($q$) instead of $b$ ($a$), and vice versa, so $a=p^N,\,b=q^N$ with rational $N:=\frac{aq}{bp}$. Write $N=u/v$ with co-prime integers $u,\,v$ with $v>0$, and write $x=z^u,\,y=z^v$ with $z=p^v=q^u\in\Bbb Q\setminus\{1\}$, so $u/v=z^{u-v}$. By the rational root theorem, $u,\,v$ are each of the form $k^{|u-v|}$ for some $k\in\Bbb Z$. Except in the case $v=1$ already covered with the version of the problem on integers, consecutive terms of this form differ by at least $2^{|u-v|}-1$, so $2^{|u-v|}-1\le|u-v|$. Hence $|u-v|=1$, and $u,\,v$ are consecutive integers, and $z=\frac{\max\{u,\,v\}}{\min\{u,\,v\}}>1$. Since $x<y$, $u<v$ and $v=u+1,\,z=1+\frac{1}{u}$. This gives the parameterization $x=\left(\frac{u+1}{u}\right)^u,\,y=\left(\frac{u+1}{u}\right)^{u+1}$, which you can verify is a solution. For example, $u=1$ gives the familiar $x=2,\,y=4$, while $u=2$ gives $x=\frac94,\,\frac{27}{8}$.
Now let's turn to the case for $\overline{\Bbb Q}$.
Since $y=x^{y/x}$, by Gelfond-Schneider $q:=y/x=x^{y/x-1}$ is rational, so $x=q^{1/(q-1)},\,y=q^{q/(q-1)}$ works for any $q\in\Bbb Q^+\setminus\{1\}$. For example, we can take $q=\frac{5}{3},\,x=\left(\frac{5}{3}\right)^{3/2},\,y=\left(\frac{5}{3}\right)^{5/2}$.
