jmc

Let $a={\log }_{2}x$ and $b={\log }_{3}y.$ Since $x>1$ and $y>1,$ $a>0$ and $b>0.$

By AM-GM, $$\begin{array}{rl}{a}^4+b^4+8& =a^4+b^4+4+4\\ & \ge 4\sqrt[4]{(a^4)(b^4)(4)(4)}\\ & =8ab.\end{array}$$Since $a^4+b^4+8=8ab,$ we have equality. Therefore, $a^4=4$ and $b^4=4.$ Then $a=\sqrt[4]{4}=\sqrt{2},$ so $$x=2^a=2^{\sqrt{2}}.$$Similarly, $b=\sqrt[4]{4}=\sqrt{2},$ so $$y=3^b=3^{\sqrt{2}}.$$Hence, $x^{\sqrt{2}}+y^{\sqrt{2}}=2^2+3^2=\boxed{13}.$