We need $3 \mid x$, so $x=3a$ and equation becomes $6a^2 - 3a = y^2$. We need $3 \mid y$, so $y=3b$ and equation becomes $2a^2 - a = 3b^2$. Rewrite as $16a^2 - 8a = 24b^2$, or $(4a-1)^2 - 6(2b)^2 = 1$. This is a Pell equation with minimal solution $(A_0,B_0) = (5,2)$ and all other solutions being $(A_{n+1},B_{n+1}) = (5A_n + 12B_n, 2A_n+5B_n)$. Indeed, $B_0$ being even, all $B_n$ will be even. However, since $A_0 \equiv 1 \pmod{4}$, we will have all $A_n \equiv 1 \pmod{4}$, thus the condition $4a-1 = A_n$ cannot be fulfilled. The equation has no positive integer solutions. To have infinitely many positive integer solutions we need take $2x^2 + 3x = 3y^2$. However, if we allow any integer solutions, the original equation also has infinitely many solutions; just take $a=(1-A_n)/4$ and $b=B_n/2$.

There is a bit of doubt surrounding this problem. It is either Show that the equation $2x^2-3x=3y^2$ has infinitely many solutions in positive integers. or Show that the equation $2x^2-3x=3y^2+y-1$ has infinitely many solutions in positive integers.

The second version leads to $3(4x-3)^2 - 2(6y+1)^2 = 1$, a Pell-like $3A^2 - 2B^2 = 1$ equation, and indeed will have infinitely many positive integer solutions.