One, Two, Three, What Do You See?

Note that $x$ has to be odd, for if $x$ is even then we would arrive at $y^{2} \equiv 3 \pmod{4}$ which is not possible.

Adding $1$ on both sides the equation becomes $y^{2}+1 = x^{3} +8 = x^{3} + 2^{3}$ .

And since $x$ was odd, lets say $x = 2n+1$ . So we have $y^{2}+1 = (2n+1)^{3}+2^{3}$ . Using the formula $x^{3}+y^{3} = (x+y) \cdot (x^{2}-xy+y^{2})$ we have $y^{2}+1 = (2n+3) \cdot (4n^{2}+4n+1 -(4n+2) + 4) = (2n+3) \cdot (4n^{2}+3)$

Now $4n^{2}+3$ is congruent to $3\pmod{4}$ so has atleast one prime divisor $p$ which is $\equiv 3 \pmod{4}$ . But this prime divisor $p$ will divide $y^{2}+1$ which is absurd, since $y^{2} \equiv -1 \pmod{p}$ is solvable iff $p \equiv 1 \pmod{4}$