Without loss of generality, let's assume that $0\le x \le y \le 10$. Because $$(x + y)^2 + (xy - 1)^2 = (x^2+1)(y^2+1)$$ is a prime, we must have $x^2+1 = 1\implies x=0$. Meanwhile, $(y^2 + 1)$ must be a prime too for $0\le y\le 10$. It is easy to check only $y=1,2,4,6,10$ meets the requirement. By symmetry, there are totally $5\times 2=\boxed{10}$ solutions.