This equation is same as $x^3 - 8 = \Phi_{17}(y)$, $\Phi_{17}(y)$ is 17th cyclotomic polynomial.
$p$ is prime number, if $p|\Phi_{17}(x)$ then $p \equiv 1 \pmod{17}$ or $p|17$. So if for arbitrary $d|\Phi_{17}(y)$ then $d \equiv 1 \pmod{17}$ or $d|17$.
$(x-2)(x^2+2x+4) = \Phi_{17}(y)$. If we have for $d = x - 2|\Phi_{17}(y)$ then $x - 2 \equiv 1 \pmod{17} \Leftrightarrow x \equiv 3 \pmod{17}$.

If we have $17|x-2$ then $x \equiv 2(17)$;
$$x^2 + 2x + 4 \equiv 9 + 6 + 4 \equiv 2(17).$$
But $x^2+2x+4$ polynomial is $\Phi_{17}(y)$ polynomial's divisor then contradict to above two cases.