Answer: C

Small Hint:

Rewrite the left side as $(x-9)^2-1$

Big Hint:

If $n=x-9,$ then the consecutive even factors $n-1$ and $n+1$ must both be powers of $2$

Solution:

Put $n=x-9.$ Then $$2^y=n^2-1=(n-1)(n+1).$$ Since the product is a power of $2,$ both factors must have no odd prime divisor. The only consecutive even integers differing by $2$ that are both signed powers of $2$ are $(-4,-2)$ and $(2,4).$ Thus $n=\pm3$ and $y=3,$ giving $x=6$ or $12.$ There are $2$ ordered pairs $(x,y).$

Therefore, the correct answer is C.