How to solve in integers the equation $2x^2+x+2x^2y-y^2+y=1$? Any hints? I could've reduced it to Pell's equation and solved it, if there wasn't $2x^2y$ part.
This is a part of a bigger problem, and that's all I have left to solve the main one.

\begin{align*}
2x^2 &+x+2x^2y-y^2+y=1
\\
\\
2(y+1)x^2 &+(1)x-(y^2-y+1)=0\\
x &= \dfrac{\sqrt{8 y^3 + 9} - 1}{4 (y + 1)}
\\ \\
y^2& - (2x^2+1)y -(2x^2+x-1)=0
\\
y &= \dfrac{2 x^2 + 1+ \sqrt{4 x^4 + 12 x^2 + 4 x - 3}}{2}
\end{align*}
Solving for $\space x\space$ suggests that
$\space y\ne-1,\space$
but solving for  $\space y\space$ and viewing a graph of this function shows that this is not necessarily so. The graph of this function suggests that solutions are not far from the origin.
A spreadsheet may not show everything but, for $\space -100 \le x \le 100,\space$ the integer solutions are
$\quad (-1,0),\space (-1,3),\space (3,-1),
\space (3,20).$

