jmc

Let $m={\log }_{b}x$ and $n={\log }_{b}y.$ Then $x=b^m$ and $y=b^n.$ Substituting into the first equation, we get $$\sqrt{b^m\cdot b^n}=b^b,$$so $b^{m+n}=b^{2b},$ which implies $m+n=2b.$

The second equation becomes $${\log }_b(b^{mn})+{\log }_b(b^{mn})=4b^4,$$so $2mn=4b^4,$ or $mn=2b^4.$

By the Trivial Inequality, $(m-n{)}^2\ge 0,$ so $m^2-2mn+n^2\ge 0,$ which implies $$m^2+2mn+n^2\ge 4mn.$$Then $(2b{)}^2\ge 8b^4,$ or $4b^2\ge 8b^4.$ Then $b^2\le \frac{1}{2},$ so the set of possible values of $b$ is $\boxed{\left(0,\frac{1}{\sqrt{2}}\right]}.$