Commutation

Since there are no solutions with $x=1$ , we can assume that $1<x<y$ .

We can write the given equation as $y(\ln{x})=x(\ln{y})$ or $\frac{\ln{y}}{y}=\frac{\ln{x}}{x}$ . Let's study the function $f(t)=\frac{\ln{t}}{t}.$ Now $f'(t)=\frac{1-\ln{t}}{(\ln{t})^2}<0$ for $x>e$ , so that $f(t)$ is decreasing for $x>e$ . If $x\geq3$ , then $f(y)<f(x)$ , so that $(x,y)$ fails to be a solution.

We still need to examine the case $x=2$ . Checking $y=3$ and $y=4$ , we find that $(x,y)=(2,4)$ is a solution, with $xy=\boxed{8}$ . For $y>4$ we have $f(y)<f(4)=f(2)$ .