How to solve the following diophantine equation?

Question

$2x^3 + x + 8 = y^2$

WolframAlpha tells me that there are no integer solutions to this equation. But how could I deduce this from observation? If so, the next step, I guess, is proving that the $LHS$ can never be a perfect square.

Answer 1

Note: The OP originally asked about the quadratic $2x^2+x+8=33$ but was later edited to ask about the cubic $2x^3+x+8$. As this answer discusses, the quadratic has integer solutions (such as $(x,y)=(8,12)$) but the cubic does not.

The quadratic which appears in the post does have integer solutions (as pointed out in the comments) but the attached WA link refers to the equation $2x^3+x+8=y^2$ which indeed has no integer solutions.

To see that the cubic equation has no integer solutions we work $\pmod 3$. Since $n^3\equiv n\pmod 3$ for all $n$, we see that the LH is always $2\pmod 3$. But $2$ isn't a quadratic residue $\pmod 3$ so we are done.