Clearly, $x>0$ and since $y^2\equiv 0,1,4\pmod 8$, $x$ must be odd. On the other hand, the equation is equivalent to $y^2+1=(x+2)(x^2-2x+4)$. If $x\equiv 1\pmod 4$, then $x+2\equiv 3\pmod 4$ and if $x\equiv -1\pmod 4$, then $x^2-2x+4\equiv 3\pmod 4$. In any case, $y^2+1$ has a positive divisor of the form $4k+3$ which is impossible.