smc

Take the base-two logarithm of both equations to get $$y{\log }_{2}x=64\text{and}({\log }_{2}y)({\log }_2{\log }_{2}x)=7.$$ Now taking the base-two logarithm of the first equation again yields $${\log }_{2}y+{\log }_2{\log }_{2}x=6.$$ It follows that the real numbers $r:={\log }_{2}y$ and $s:={\log }_2{\log }_{2}x$ satisfy $r+s=6$ and $rs=7$. Solving this system yields $$\{{\log }_{2}y,{\log }_2{\log }_{2}x\}\in \{3-\sqrt{2},3+\sqrt{2}\}.$$ Thus the largest possible value of ${\log }_{2}y$ is $3+\sqrt{2}⟹\boxed{3+\sqrt{2}}$.