Can you solve?

$\begin{aligned} 2^{x^y} & = x \\ x^y \log 2 & = \log x \\ x^y & = \frac {\log x}{\log 2} \end{aligned}$

Since the LHS $x^y$ is an integer, the RHS must also be an integer. $\implies x = 2^k$ , where $k \in \mathbb N$ .

$\begin{aligned} \implies 2^{ky} & = \frac {\log 2^k}{\log 2} = k \\ ky \log 2 & = \log k \\ ky & = \frac {\log k}{\log 2} & \small \color{#3D99F6} \text{Again } k = 2^n \text{ a power of }2 \\ \implies y & = \frac n{2^n} < 1 & \small \color{#3D99F6} \text{For }n \in \mathbb N \end{aligned}$

Therefore, $y=0$ and $xy = \boxed{0}$ .

For $x = 2$ and $y= 0,$

$\large 2^{x^y} = x$

$\large \Rightarrow 2^{2^0} = 2.$

Hence $xy= 2 \times 0 = \boxed{0}$

For y>=1, 2^x^y is always greater than x.

For y<=-1, 2^x^y is always irrational, unless x=0.

So, either x=0 or y=0. That alone is enough to know the answer is 0, if there is one. But to find the actual solution:

Trying x=0, we get 1=0, so that's no good.

Trying y=0, we get 2^1=x. So the solution is x=2, y=0.