Solution:

The answer is (E). Let us rewrite the expression as $x y - 4 x = 4 y^{2}$, that is, $x(y-4) = 4 y^{2}$. $y = 4$ is not a solution, so we can divide by $y-4$ and obtain the equation
$$x = \frac{4 y^{2}}{y-4}.$$
Observing that $y^{2} = (y+4)(y-4) + 16$ we can further rewrite the equation in the form
$$x = 4(y+4) + \frac{64}{y-4},$$
so the solutions of the original equation correspond to the integer values of $y$ for which $\frac{64}{y-4}$ is an integer (indeed, for such values of $y$ there exists a unique choice of $x$, given by the previous expression, that provides a solution of the original equation).
It follows that the solutions sought are as many as the divisors (positive and negative) of $64$: these are all of the form $\pm 2^{i}$ with $i=0,1, \ldots, 6$, hence there are exactly $2 \cdot 7 = 14$.