Showing that the only integer solutions to $y^3=4x^2+47$ are $x=250,-250$ We’ve computed that the class number for $\mathbf Q(\sqrt{-47})$ is $5$. From my attempt at this question, we have the equation of ideals
$$(y)^3=(2x+\sqrt{-47})(2x-\sqrt{-47}).$$
Next we show that the ideals on the RHS are coprime. If an ideal $\mathfrak a$ divides both ideals on the RHS, then it divides $(2\sqrt{-47}) = (2)(\sqrt{-47})$. Also the ideal norm ${\rm N}(\mathfrak a)$ divides $y^6$, which is odd, so $\mathfrak a$ is relatively prime to $(2)$. Thus  $\mathfrak a \mid (\sqrt{-47})$, so $\mathfrak a$ is $(1)$ or $(\sqrt{-47})$. If $\mathfrak a = (\sqrt{-47})$ then $47 \mid x$, so also $47 \mid y$ from the equation $y^3 = 4x^2 + 47$, but then $47 = y^3 - 4x^2$ is divisible by $47^2$ and we have a contradiction. So $\mathfrak a = (1)$: $(2x+\sqrt{-47})$ and $(2x-\sqrt{-47})$ are relatively prime.
We can also try to use the class number by noting that $(2x\pm\sqrt{-47})^2$ are both principal ideals and also have sixth roots, but that’s pretty much what we already know…
Then, we take an attempt at the sixth roots—if there exists such sixth root $a+b(\frac{1+\sqrt{-47}}{2})$, then we must have, after computation, $$3a^2b+3ab^2-11b^3=2$$ Hence we must have $b=1,2,-1,-2$. The only case where this has a solution is when $(a,b)=(-5,2),(3,2)$.
 A: Here’s a complete solution if anybody still needs it, or needs an example of solving Diophantine Equations with class numbers.
We have the equation of ideals $(y)^3 = (2x+\sqrt{-47})(2x-\sqrt{-47})$ where the ideals on the right side are coprime by an argument in the post, so by unique factorization of ideals they are each cubes of ideals. Write $(2x+\sqrt{-47}) = \mathfrak b^3$. Since the class number $5$ is relatively prime to $3$, $\mathfrak b$ has to be principal: in the ideal class group, the ideal class of $\mathfrak b$ has trivial $3$rd power and $5$th power, so it is has trivial $1$st power.
Thus
$$
(2x+\sqrt{-47}) = (\alpha)^3 = (\alpha^3) 
$$
for some $\alpha \in \mathbf Z[(1+\sqrt{-47})/2]$, so
$2x+\sqrt{-47}$ is a unit multiple of $\alpha^3$.
The units in this ring of integers are $\pm 1$, and both of them are cubes, so we can absorb a unit factor into $\alpha$: without loss of generality
$$
2x + \sqrt{-47} = \alpha^3.
$$
Now we solve for the two coefficients for $\alpha$, and one is easy to solve in integers, and we find that $2x=500,-500$, so if there is a solution then $x$ must be $250$ or $-250$.
