First of all, let us simplify the equation by replacing things that the attacker can compute with known constants. We come up with:
$$a \cdot b^x = y$$
where the attacker knows $a$ (which is $e(g,h)^k$) and $b$ (which is $e(g, h)$, which he can compute, as he knows $g, h$), and the attacker solves for $x, y$.
If it is sufficient for an attacker to find a solution, he can do that easily; he can select a random $x$, and solve it for $y$; there's a solution.
If he needs to find "the correct" solution (that is, the one possible $x, y$ pair that satisfies additional criteria), he doesn't have enough information to pick out which one it is. He might be able to if he knows what that additional criteria is; it would depend on what that looks like.