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?
 A: You have that $3y-2$ divides $y^2$.
Notice that $(3y-2,y^2)| (3y-2,y)^2$.
On the other hand, if $d$ divides $3y-2$ and $y$ we have $d|2$. So $(3y-2,y^2)| 4$.
Therefore $3y-2|4$
Therefore $y=0,1$ or $2$.
If $y=0$ we have $x=0$.
If $y=1$ we have $x=\pm 1$
If $y=2$ we have $x=\pm 1$
A: 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.
A: We note that $x^2\mid 3x^2y$, so $x^2\mid 2x^2+y^2$, then $x^2\mid y^2$, which implies that $x\mid y$. Let's call $y=kx$, if we replace this into your equation $(1)$ we get $k^2x^2=x^2(3kx-2)$, then $k^2=3kx-2$, which lead us to the quadratic equation $k^2-3xk+2=0$. Using the formula for quadratic equations we deduce that $$k=\frac{3x\pm \sqrt{9x^2-8}}{2}.$$
So in order to have $k\in\mathbb{Z}$ we must have $9x^2-8=z^2$, for some $z\in  \mathbb{Z}$, so $(3x+z)(3x-z)=8$, therefore $3x+z$ and $3x-z$ are both divisors of $8$ with the same parity. Hence $3x+z=\pm 4$ and $3x-z=\pm 2$, or $3x+z=\pm 2$ and $3x-z=\pm 4$. 
For example if $3x+z=4$ and $3x-z=2$, then $x=1$ and $z=1$, which gives us $k=1$ or $k=2$, and thus $y=1$ or $y=2$. The other cases are similar.
A: $x=0\iff y=0.$ And $x=y=0$ is a solution.
If $x\ne 0\ne y$ then $$2x^2+y^2=3x^2y\implies x^2(2-3y^2)=-y^2\ne 0.$$  This implies $x^2$ divides $y^2.$ Now $y^2/x^2$ is an integer for non-zero integers $x,y$ only if  $x$ divides $y$.  So we have $$y=zx$$ with  integer $z.$
So $2x^2+z^2x^2=3x^3z$. Since $x\ne 0$ we have $2+z^2=3xz.$... So $$z(z-3x)=-2$$ which implies that $z$ divides $2$.
Putting $y=zx$ with $z\in \{\pm 1 ,\pm 2\}$ into the original equation gives us the non-zero solutions :$$(y=1, x=\pm 1), (y=2,x=\pm 1).$$ For example if $z=1$ then $y=x$ and $3x^2=2x^2+y^2=3x^2y=3x^3\ne 0$ implies $1=x=y.$
