smc

Let ${\log }_{2}x={\log }_{y}16=k$, so that $2^k=x$ and $y^k=16⟹y=2^{\frac{4}{k}}$. Then we have $(2^k)(2^{\frac{4}{k}})=2^{k+\frac{4}{k}}=2^6$. We therefore have $k+\frac{4}{k}=6$, and deduce $k^2-6k+4=0$. The solutions to this are $k=3\pm \sqrt{5}$. To solve the problem, we now find $$\begin{array}{rl}({\log }_2\frac{x}{y}{)}^2& =({\log }_{2}x-{\log }_{2}y{)}^2\\ & =(k-\frac{4}{k}{)}^2=(3\pm \sqrt{5}-\frac{4}{3\pm \sqrt{5}}{)}^2\\ & =(3\pm \sqrt{5}-[3\mp \sqrt{5}]{)}^2\\ & =(3\pm \sqrt{5}-3\pm \sqrt{5}{)}^2\\ & =(\pm 2\sqrt{5}{)}^2\\ & =\boxed{20}.\end{array}$$