Shouldn't be integers? Clearly there are solutions if both are real... E.g. $x=1, y=\sqrt{3}$

Indeed, you must assume $x,y\in\mathbb{N}$. With this assumption, Note that since $4x^2+1=(2x)^2+1$ is odd, $y$ must be odd, as well. Hence, $y^2+2\equiv 3\pmod{4}$. In particular, it has a prime divisor $p$ for which $p\equiv 3\pmod{4}$. If $\frac{4x^2+1}{y^2+2}$ had been an integer, then we would have, $(2x)^2\equiv -1\pmod{p}$, which is a contradiction (since $-1$ is not a quadratic residue modulo $4k+3$).