<s>[INST] Find all integer solutions $(x,y)$ such that $2x^2 + y^2 = 3x^2y$ 
Find all integer solutions $(x,y)$ such that $2x^2 + y^2 = 3x^2y$.

We can rearrange the given equation to $$y^2 = x^2(3y-2)\tag1$$ Thus $3y-2$ must be a perfect square and so $3y-2 = k^2$.
How can we continue?
 [/INST] I. The method described by the OP is a bit round-about. To continue, solve for $y$ in $3y-2=k^2$ then substitute into $y^2-x^2(3y-2)=0$. We get,
$$(2 + k^2 - 3 k x) (2 + k^2 + 3 k x)=0$$
Solve for $k$,
$$k =\frac{3x\pm\color{blue}{\sqrt{-8+9x^2}}}{2}$$
Thus one needs to solve $-8=z^2-9x^2$ in the integers. It is a Pell-like equation with a very small number of solutions and easily done.
II. Or, more quickly, just solve for $y$ in the original equation,
$$2x^2 + y^2 = 3x^2y$$
to get,
$$y =\frac{3x^2\pm x\color{blue}{\sqrt{-8+9x^2}}}{2}$$
which gives the same easy condition to solve.
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