I tried to find the solution of this equation, or conclude there are none. This i what i found out:
I noticed that $x \neq -1$ mod $3$.
We can write $y^2 + 4 = x^3 - 8 = (x-2)(x^2+2x+4)$
I tried to find a prime divider from the righthand side so i could use legendre on the left hand side, but couldnt solve it. Can someone help me?
On
A general problem considering the equations of the form $$ y^2 = x^3 + k$$ are known as Bachet equations. There is a known theorem as follows (which could be found in Richard Mollin's Algebraic Number Theory: theorem 4.2) :
Let $F=\mathbb{Q}(\sqrt{k})$ be a complex quadratic field with radicand $k< -1$ such that $k \neq 1 \pmod 4,$ and $h_{\mathfrak{D}_F} \neq 0\pmod 3.$ Then there are no solutions to the Batchet equation in integers $x,y$ except in the following cases: there exists an integer $u$ such that $$(k,x,y) = (\pm 1-3u^2, 4u^2 \mp 1, \epsilon \cdot u(3 \mp 8u^2) ),$$
where the $\pm$ signs correspond to the $\mp$ signs and $\epsilon = \pm 1$ is allowed in either case.
In this case, $k=-12<-1$ and $k\equiv0\pmod 4$, $h_{\mathfrak{D}_F}=1$, however, $12\pm 1\neq0\pmod 3$, hence there is no integer solutions.
As you've noticed, $y^2+4=(x-2)\left(x^2+2x+4\right)$.
If $x$ is even, then let $x=2k$. Then $y=2m$, so $m^2=2k^3-3$, impossible (use mod $8$, as has been suggested in the comments. $2k^3\equiv \{0,2,6\}\pmod{8}$, but then $m^2\equiv \{5,7,3\}\pmod{8}$, contradiction, because $5,7,3$ are not quadratic residues mod $8$).
If $x$ is odd, then: let $p$ be a prime divisor of either $x-2$ or $x^2+2x+4$. Then $p$ is odd and $p\mid y^2+4$, so $\left(y2^{-1}\right)^2\equiv -1\pmod{p}$, so $p=4t+1$ (by Quadratic Reciprocity).
Therefore all the prime divisors of $x-2$, $x^2+2x+4$ are of the form $4t+1$. Also $x^2+2x+4=(x+1)^2+3>0$ and $y^2+4>0$, so $x-2>0$, so $x-2\equiv 1\pmod{4}$ and $x^2+2x+4\equiv 1\pmod{4}$. The first congruence gives $x\equiv 3\pmod{4}$, but $3^2+2\cdot 3+4\not\equiv 1\pmod{4}$, contradiction.