$(a, b) = (4, 2)$。

Let $d$ be the greatest common divisor of $a$ and $b$, $a = dp$, $b = dq$, where $p, q$ are coprime positive integers and $p > q$. Substitute into the original equation:
$$(d(p-q))^{d^2pq} = (dp)^{dq} \cdot (dq)^{dp} \quad \Leftrightarrow \quad (d(p-q))^{dpq} = (dp)^q \cdot (dq)^p \\ \Leftrightarrow d^{dpq}(p-q)^{dpq} = d^{p+q}p^q q^p.$$

i. First, prove $p+q < dpq$. Otherwise, $p+q \ge dpq$, rearrange to get $(p-q)^{dpq} = d^{p+q-dpq}p^q q^p$, so $p, q$ are both factors of $(p-q)^{dpq}$. But $(p-q,p) = (p-q,q) = (p,q)=1$, so $p=q=1$, which is not valid.

Since $p+q < dpq$, rearrange to get $d^{dpq-p-q}(p-q)^{dpq} = p^q q^p$, so $p-q$ divides $p^q q^p$. But $(p-q,p) = (p-q,q) = 1$, so $p-q=1$, i.e., $p=q+1$. Substitute back to get
$$d^{dpq-p-q} = p^q q^{q+1}. \qquad (1)$$

ii. Since $p, q$ are coprime, $d$ can be written as $d = m \times n$, where $m, p$ share the same prime factors, $n, q$ share the same prime factors. Thus $(m, n) = 1$, from (1) we get
$$m^{dpq-p-q} = p^q = (q+1)^q. \qquad (2)$$
Since the exponents $dpq-p-q$ and $q$ are coprime, $m$ must be a $q$th power of some integer. Let $m = t^q$, then $t^{dpq-p-q} = q+1$. Since $q+1 > 1$, $t > 1$.

iii. If $q=2$, then $t^{6d-5} = 3$, so $d=1, t=3, q=2, p=3$, i.e., $(a,b)=(3,2)$. But $(a,b)=(3,2)$ does not satisfy the original equation, discard.

If $q \ge 3$, then $dpq-p-q = dq(q+1) - (2q+1) \ge 3(q+1) - (2q+1) = q+2$. Then $q+1 = t^{dpq-p-q} \ge t^{q+2} \ge 2^{q+2} \ge q+2$ (by binomial theorem), which is not valid.

So $q=1, p=2$. From $t^{2d-3} = 2$, we get $t=2, d=2$. Thus $(a,b) = (4,2)$. Substitute back into the original equation and it holds.

Therefore, $(a,b) = (4,2)$ is the unique solution.