jmc

We begin by rewriting the equation as $2x^2{y}^3-3x^2+4y^3=149$. We can then use Simon's Favorite Factoring Trick and subtract 6 from both sides of the equation to get $2x^2{y}^3-3x^2+4y^3-6=143$. This can be factored into $$(x^2+2)(2y^3-3)=143$$Since we know that the prime factorization of $143=11\cdot 13$, we have that $2y^3-3$ must equal $\pm 1,\pm 11,\pm 13$ or $\pm 143$. The only possible values of $y$ are $1$ and $2.$ For $y=1,$ there are no solutions. For $y=2,$ we have $x=3.$ Therefore, $x+y=\boxed{5}$.