The equality implies $x-8$ and $x-10$ are both powers of two; since they differ by two, it must be the case that $(x-8,x-10) = (4,2)$ or $(x-8,x-10) = (-2,-4)$.  (Note that $(1,-1)$ is not allowed because then the product is negative.) These yield $(x,y) = (12,3)$ or $(x,y) = (6,3)$, for a total of $\boxed{2\textbf{ (C)}}$ solutions.
If both $x$ and $y$ are both integers, how many pairs of solutions are there of the equation $(x-8)(x-10) = 2^y$?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ \text{more than 3}$