Number of solutions!

This equation can be rewritten as $y = \dfrac{x^{2} + 1}{3}.$ So for $y$ to be an integer, we will require that

$x^{2} + 1 \equiv 0 \pmod {3} \Longrightarrow x^{2} \equiv 2 \pmod{3}.$

But if $x \equiv 0,1,2$ modulo $3$ then $x^{2} \equiv 0,1,1$ modulo $3,$ respectively, which means that $2$ can never be a quadratic residue modulo $3,$ and hence no integer $x$ exists that can make $y$ an integer.

Thus there are $\boxed{0}$ solutions $(x,y).$