$x^3+48=y^4$ does not have integer (?) solutions How does one find all positive integer solutions to the equation $x^3+48=y^4$?

I'm posting this as a community wiki "answer" because I'm not sure it works and the explanation wouldn't show up very well in the comments section I guess. If we factor the equation in $\mathbb{Z}[\sqrt{3}]$ we get $(y^2 - 4\sqrt{3})(y^2 + 4\sqrt{3}) = x^3$. Thus if $y^2 - 4 \sqrt{3} = (a + b\sqrt{3})^3$ for some $a + b\sqrt{3} \in \mathbb{Z}[\sqrt{3}]$ then we get
$$y^2 - 4 \sqrt{3} = a^3 + 3a^2b\sqrt{3} + 9ab^2 + 3b^3 \sqrt{3} $$
so that we must have $3a^2 b + 3b^3 = -4$ and since $3$ does not divide $4$ we get a contradiction. The problem is that I need to justify somehow that $y^2 - 4 \sqrt{3} = (a + b\sqrt{3})^3$ and although I know that $\mathbb{Z}[\sqrt{3}]$ is a unique factorization domain, I would need the two factors $y^2 - 4 \sqrt{3}$ and $y^2 + 4 \sqrt{3}$ to be coprime I think, and I can't seem to justify it or disprove it.
Edit (Alex):
Suppose that we have an integer solution to $x^3 = y^4 - 48$.
Claim: The two factors $y^2\pm 4\sqrt{3}$ are coprime in $\mathbb{Z}[\sqrt{3}]$. Indeed, let $\alpha = v+\sqrt{3}w$ divide both factors. Then, $\alpha$ also divides their difference, which is $8\sqrt{3}$. Without loss of generality assume that $\alpha$ is irreducible. Then, $\alpha$ divides 2 or is (up to a unit) equal to $\sqrt{3}$. Since $\alpha$ also divides $x^3\in \mathbb{Z}$, we deduce that $x^3$ is divisible by either 2 or 3 (or both). This implies that $y$ must also be divisible by either 2 or 3. But then, $x^3$ is also divisible by either 8 or 27 (or both), since it is a cube, and $y^4$ is divisible by 16 or 81. Now, the latter is impossible, since $y^4 - 48 \equiv -3\neq 0\pmod{9}$. The former is also impossible, since it would imply that $x^3=y^4 - 48$ is divisible by 16, so $x$ is divisible by 4, so $x^3$ is divisible by 64 and you would obtain a contradiction modulo 64 as above.
This implies that $y^2+4\sqrt{3}$ is a cube times a unit in $\mathbb{Z}[\sqrt{3}]$, $y^2+4\sqrt{3}=u\cdot(a+\sqrt{3}b)^3$. The units in this ring form an abelian group with torsion $\pm 1$ and with free part of rank one, generated by $2+\sqrt{3}$. The torsion are all cubes, so we can incorporate them into the bracket. Of course, cubes of a non-torsion unit can also be incorporated into the bracket, so it suffices to show that $y^2+4\sqrt{3}\neq (2+\sqrt{3})^{\delta}\cdot(a+\sqrt{3}b)^3$ for $\delta\in{0,1,2}$. This is done through case by case analysis, Adrián has already done the $\delta=0$ case.
