One of my motivations for studying this problem is mathematical curiosity, as I am aware that it has the solution $2$. I am intrigued to determine whether 2 is the sole solution or if there are others.
The equation also exhibits an intriguing pattern:
...$=2$^^^$2=2$^^$2=2^2=2⋅2=2+2=1+1+1+1.$
In my opinion, this equation is worth solving.
(Partial answer)
The whole set of solutions might be hard to determine in the present case. By inspection, we can find some of them, such as $x = y = 1$, $x = y = 2$ or $x = 0$ with $y > 0$ for instance. Provided suitable domains for $x$ and $y$, the Lambert W function permits to find a solution curve too, since your equation can be rewritten as $\frac{1}{x} = ye^{-y\ln x}$, hence $y = \frac{W\left(-\frac{\ln x}{x}\right)}{-\ln x}$.
Now, given that you are ultimately interested by the fact that $2+2 = 2\cdot2 = 2^2$, you may look at hyperoperations, because it is known that $H_n(2,2) = 4 \;\,\forall n \ge 1$ actually.