Solving the Diophantine equation $y^3 = 4x^2+4x+ 5$ for $x,y \in \mathbb{Z}$ 
I want to solve the Diophantine equation $y^3 = 4x^2+4x+ 5$ for $x,y \in \mathbb{Z}$.

The right hand side factors as $(2x+1-2i)(2x+1+2i)$. 
Am I right that such a factorization can be found using the quadratic formula?
Now I think that the equation can be solved by performing a descent as $\mathbb{Z}[i]$ is a UFD. But for doing that we need that the above two factors are coprime. 
Are the above two factors coprime, and how do I show that?
 A: Let $x,y\in\mathbb{Z}$ be such that $y^3=4x^2+4x+5=(2x+1-2\text{i})(2x+1+2\text{i})$.  Let $d\in\mathbb{Z}[\text{i}]$ be a greatest common divisor of $2x+1-2\text{i}$ and $2x+1+2\text{i}$. Thus, $d$ divides the difference $4\text{i}$, but the norm of $d$ must be odd (as the norm of $d$ divides the odd integer $4x^2+4x+5$).  This shows that the norm of $d$ is an odd number dividing the norm of $4\text{i}$, which is $16$.  Hence, $d$ must have a unit norm, whence it is a unit.  Ergo, $\gcd(2x+1-2\text{i},2x+1+2\text{i})=1$ in $\mathbb{Z}[\text{i}]$.
Since $\gcd(2x+1-2\text{i},2x+1+2\text{i})=1$ and the units of $\mathbb{Z}[\text{i}]$ are perfect cubes in $\mathbb{Z}[\text{i}]$, both $2x+1-2\text{i}$ and $2x+1+2\text{i}$ must be perfect cubes in $\mathbb{Z}[\text{i}]$.  That is, for some $u,v\in\mathbb{Z}$, 
$$2x+1+2\text{i}=(u+v\text{i})^3\,,$$
which implies
$$2=(3u^2-v^2)v\,.$$
Consequently, $v\in\{\pm1,\pm2\}$.  It is easy to see that only $(u,v)=(\pm1,1)$ and $(u,v)=(\pm1,-2)$ works.  This gives $$2x+1=u(u^2-3v^2)\in\{\pm2,\pm11\}\,.$$
That is, $x=5$ or $x=-6$, leading to the two solutions $(x,y)=(5,5)$ and $(x,y)=(-6,5)$.
