Well, after a couple of days thinking I think the diophantine equation can be solved in a different way. First of all, asumme that $y=0$, so the equation becomes $8x^6=0$, then $x=0$. Now, let's consider $y\neq 0$, we note that the LHS is divisible by $y$ and thus the RHS also has to be divisible by $y$, so we get that $y\mid 8x^6$. The idea is to prove that $\gcd(x,y)=1$. We can write the equation in the form $$(2x^3+y^2+y)(4x^3-y)=4x^3y^2.$$
Let's suppose $\gcd(x,y)=d>1$, so there exists a prime number $p$ such that $p\mid x$ and $p\mid y$. Set $v_{p}(x)=\alpha>0$ and $v_{p}(y)=\beta>0$. First, let's assume that $p$ is odd, then $v_{p}(4x^3y^2)=3\alpha+2\beta$.
On the other hand, $v_{p}((2x^3+y^2+y)(4x^3-y))=v_{p}(2x^3+y^2+y)+v_{p}(4x^3-y)$, but $v_{p}(2x^3+y^2+y)=\min\{v_{p}(2x^3), v_{p}(y^2), v_{p}(y)\}=3\alpha$ or $\beta$ if $3\alpha\le \beta$ or if $3\alpha>\beta$, respectively. Also $v_{p}(4x^3-y)=\min\{v_{p}(4x^3), v_{p}(y)\}=3\alpha$ or $\beta$ if $3\alpha\le \beta$ or if $3\alpha>\beta$, respectively. In summary, we deduce that
$$v_{p}((2x^3+y^2+y)(4x^3-y))=
\begin{cases}
6\alpha &\text{if } 3\alpha\le \beta \\
2\beta &\text{if } 3\alpha>\beta
\end{cases}
$$
So we conclude that $6\alpha=3\alpha+2\beta$ or $2\beta=3\alpha+2\beta$. In the first case we get $\beta\le 0$, and in the second case we get $\alpha=0$, contradiction in both cases. Now, Let's suppose $p=2$. In this case we get $v_{p}(4x^3y^2)=3\alpha+2\beta+2$, and after an analogous reasoning we deduce that
$$v_{p}((2x^3+y^2+y)(4x^3-y))=
\begin{cases}
6\alpha+3 &\text{if } 3\alpha+1< \beta \\
3\alpha+\beta+1 &\text{if }\, 3\alpha+1=\beta \\
2\beta &\text{if } 3\alpha+1>\beta
\end{cases}
$$
So we have either $6\alpha+3=3\alpha+2\beta+2$ or $3\alpha+\beta+1=3\alpha+2\beta+2$ or $2\beta=3\alpha+2\beta+2$. In any case we arrive to a contradiction. In conclusion such $p$ doesn't exists and therefore $\gcd(x,y)=1$. Using this and the fact that $y\mid 8x^6$ we get that $y\mid 8$. Therefore, the possible values for $y$ are $\{\pm1, \pm2, \pm4, \pm8\}$.
A routine check using the possible values of $y$ gives us solutions only for $y=-1$ and $y=2$. In the first case we get $x=0$, and in the second case we get $x=1$. Hence, all the integer solutions for the equation are: $$(x,y)=(0,0), (0,-1), (1,2).$$
