How many solutions exist?

Taking mod 4, we have $y^2 \equiv x^3 - 1 \! \pmod{4},$ so $x$ must be odd i.e. $x = 2a + 1.$ Now, rewrite the equation as $y^2 + 1 = x^3 + 2^3 = (x + 2)(x^2 - 2x + 4).$ Note that

$x^2 - 2x + 4 \equiv x(x - 2) \equiv (2a + 1)(2a - 1) \equiv 4a^2 - 1 \equiv 3 \! \! \! \! \pmod{4}.$

Therefore, $x^2 - 2x + 4$ has a prime factor $p$ of the form $4k + 3.$ This prime must also divide $y^2 + 1,$ so that $y^2 \equiv -1 \! \pmod{p}.$ However,

$\left ( \dfrac{-1}{p} \right ) = (-1)^{\frac{p - 1}{2}} = (-1)^{\frac{4k + 3 - 1}{2}} = (-1)^{2k + 1} = -1,$

so no $y$ exist such that $y^2 \equiv -1 \! \pmod{p}.$ We conclude that there are $\boxed{0}$ solutions to the equation.