Let $x = dp$, $y = dq$, where $d = (x, y)$ is the greatest common divisor of $x, y$, $p, q \in \mathbb{N}$, $(p, q) = 1$, $p > q$, then
$$(d(p-q))^{d^2pq} = (dp)^{dq}(dq)^{dp} \Leftrightarrow (d(p-q))^{dpq} = (dp)^q(dq)^p \Leftrightarrow d^{dpq}(p-q)^{dpq} = d^{p+q}p^q q^p.$$

We want to prove: $p+q < dpq$,
Suppose $p+q \ge dpq$, then $(p-q)^{dpq} = d^{p+q-dpq}p^q q^p \Rightarrow p \mid (p-q)^{dpq}$ and $q \mid (p-q)^{dpq}$, but $(p-q, p) = (p-q, q) = (p, q) = 1 \Rightarrow p=1, q=1$ (contradiction) therefore $p+q < dpq$, so $d^{dpq-p-q}(p-q)^{dpq} = p^q q^p \Rightarrow (p-q) \mid p^q q^p$, and also $(p-q, p) = (p-q, q) = (p, q) = 1 \Rightarrow p-q=1$, that is $p=q+1$, therefore
$$d^{dpq-p-q} = p^q q^{q+1}, \qquad (1)$$
so $p, q$ are divisors of $d$. Since $(p, q)=1$, $d$ can be written as $d=s \times t$, where the prime factors of $t$ are only $p$, the prime factors of $s$ are only $q$, and $(s, t)=1$. From equation (1) we get:
$$t^{dpq-p-q} = p^q = (q+1)^q. \qquad (2)$$
Since $dpq-p-q$ is coprime to $q$, from (2) we know $t$ must be a $p$-th power of some natural number, let $t = t_1^q$. Then
$$t_1^{dpq-p-q} = q+1 \Leftrightarrow t_1^{dq(q+1)-(2q+1)} = q+1, \text{ since } q+1 > 1, \text{ hence } t_1 > 1. \text{ If } q \ge 3, \text{ then}$$
$$dq(q+1)-(2q+1) \ge 3(q+1)-(2q+1) = q+2 \Rightarrow t_1^{dpq-p-q} \ge 2^{q+2} > q+1 \text{ (contradiction)}$$
If $q=2$, then $t_1^{6d-5}=3 \Rightarrow t_1=3$, $d=1$ and $x=d(q+1)=3$, $y=dq=2$ (contradiction). Therefore
$q=1$, then $t_1^{2d-3}=2 \Rightarrow t_1=2$, $d=2 \Rightarrow x=4, y=2$.