smc

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 $\boxed{2}$ solutions.