The equation is equivalent to the following one

$$\begin{aligned} & \left(9 y^{2}+6 y+1\right) x^{2}+\left(9 y^{2}+18 y+5\right) x+2 y^{2}+7 y+6=0 \\ & \Leftrightarrow(3 y+1)^{2}\left(x^{2}+x\right)+4(3 y+1) x+2 y^{2}+7 y+6=0 \end{aligned}$$

Therefore $3 y+1$ must divide $2 y^{2}+7 y+6$ and so it must also divide

$$9\left(2 y^{2}+7 y+6\right)=18 y^{2}+63 y+54=2(3 y+1)^{2}+17(3 y+1)+35$$

from which it follows that it must divide 35 as well. Since $3 y+1 \in \mathbb{Z}$ we conclude that $y \in\{0,-2,2,-12\}$ and it is easy now to get all the solutions $(-2,0),(-3,0),(0,-2),(-1,2)$.