If we add these two equalities, we have equality $4 + x = 2a^4$, if we subtract, $x - 4 = \frac{2}{a^2}$. Then we have, that $(x+4)(x-4)^2 = \frac{4}{a^4} \cdot 2a^4 = 8$. As we are interested only in integer $x$, each of expressions $x+4$ and $x-4$ has to be integer. As these two numbers are of the same parity and $(x-4)^2$ is a square of an integer, equalities $x+4=2$ and $(x-4)^2 = 4$ must be true. But it is impossible for both to be true at the same time. Thus, there is no such value $x$.