The considered equation is equivalent to
$$(2y + x)^2 = 4x^3 - 27x^2 + 44x - 12 = (x - 2)(4x^2 - 19x + 6) = (x - 2)((x - 2)(4x - 11) - 16).$$
The expression above must be a perfect square. Therefore we have either $x = 2$ (and $y = -1$), or $(x - 2) = ks^2$, where $k \in \{-2, -1, 1, 2\}$ and $s \in \mathbb{N}$; indeed, if for some prime $p$ and nonnegative integer $m$ the number $p^{2m+1}$ divides $(x-2)$ but $p^{2m+2}$ does not, then we have $p \mid (x-2)(4x-11) - 16$, so $p \mid 16$ and $p = 2$.
We will consider three cases separately.

The case of $k = \pm 2$. Then we have $4x^2 - 19x + 6 = \pm 2u^2$ for some integer $u$, or, equivalently,
$$(8x - 19)^2 - 265 = \pm 32u^2,$$
a contradiction modulo $5$.

The case of $k = 1$. Then $4x^2 - 19x + 6 = u^2$ for some integer $u$, which leads to
$$265 = (8x - 19)^2 - 16u^2 = (8x - 19 - 4u)(8x - 19 + 4u).$$
We easily check that $x = 6$ is the only solution of this equation (we simply consider all possible decompositions: $265 = 1 \cdot 265 = 5 \cdot 53 = \dots$, etc, and take into account the fact that $x - 2 = s^2$). Therefore, we obtain two solutions of the original equation: $(x, y) \in \{(6, 3), (6, -9)\}$.

The case of $k = -1$. As before, we have $4x^2 - 19x + 6 = -u^2$, which is equivalent to
$$265 = (8x - 19)^2 + (4u)^2$$
and we check all possibilities with $u \le 4$: for $u = 0, 1, 2$ there are no solutions. If $u = 3$, then we obtain $(8x - 19)^2 = 121 = 11^2$, which leads to $x = 1$, which gives two solutions: $(x, y) \in \{(1, 1), (1, -2)\}$. Finally, for $u = 4$ we arrive at $(8x - 19)^2 = 9 = 3^2$ and $x = 2$, which gives $(x, y) = (2, -1)$.

Therefore, the set of solutions is as follows:
$$$$
\{(6, 3), (6, -9), (1, 1), (1, -2), (2, -1)\}.