Solution:

Answer: $y = \sqrt{2}$

We note that $\left(a \otimes b^{k}\right) = (a \otimes b)^{k}$ for all positive integers $k$. Then for all rational numbers $\frac{p}{q}$ we have $a \otimes b^{\frac{p}{q}} = \left(a \otimes b^{\frac{1}{q}}\right)^{p} = (a \otimes b)^{\frac{p}{q}}$. So by continuity, for all real numbers $a, b$, it follows that $2^{a} \otimes 2^{b} = (2 \otimes 2)^{a b} = 4^{a b}$. Therefore given positive reals $x, y$, we have

$x \otimes y = 2^{\log_{2}(x)} \otimes 2^{\log_{2}(y)} = 4^{\log_{2}(x) \log_{2}(y)}$.

If $x = 4^{\log_{2}(x) \log_{2}(y)} = 2^{2 \log_{2}(x) \log_{2}(y)}$ then $\log_{2}(x) = 2 \log_{2}(x) \log_{2}(y)$ and $1 = 2 \log_{2}(y) = \log_{2}\left(y^{2}\right)$. Thus $y = \sqrt{2}$ regardless of $x$.