Czech-Polish-Slovak Match

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 $(x-2)=k{s}^2$, 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)\}.