For $x$ to be in the domain, there must exist an integer $y$ such that $x^2 + y^2 = 9$. If $x=0$, $y^2=9$ ($y=\pm3$). If $x=3$, $y^2=0$ ($y=0$). If $x=-3$, $y^2=0$ ($y=0$). If $x=2$, $y^2 = 9 - 4 = 5$. Since $\sqrt{5}$ is not an integer, $x=2$ cannot be in the domain of a relation defined over integers $\mathbb{Z}$. Thus, 2 is the correct answer. All other options result in integer values for $y$.