Solution:

Let $a = \log_2 x$ and $b = \log_2 y.$ Taking $\log_2$ of $x^y = 2^{64}$ gives $y \log_2 x = 64,$ i.e. $a \cdot 2^b = 2^6.$

Taking $\log_2$ of the second equation gives $b \log_2 a = 7,$ so $a = 2^{7/b}.$ Substituting, $2^{7/b} \cdot 2^b = 2^6,$ so $b + \dfrac7b = 6,$ i.e. $b^2 - 6b + 7 = 0.$

Thus $b = 3 \pm \sqrt2,$ and the greatest value of $\log_2 y$ is $3 + \sqrt2.$

Thus, the correct answer is C.