Difference in choosing some generator of principal ideal for solving $x^2+d=y^3$ So in solving $E : x^2 + 19 = y^3$ in the integers I determined that $(x + \sqrt{-19}) = I^3$ for some ideal in $$\mathscr{O}_K = \mathbb{Z}\left[\frac{1 + \sqrt{-19}}{2}\right]$$ and since the class number of $K$ is not $3$, that $I$ must be principal. 
So what I did was $$\left(a + b \frac{1 + \sqrt{-19}}{2}\right)^3 = x + \sqrt{-19},$$ but the equation I ended up with doesn't have solutions: $b(3a^2+3ab-4b^2)=1$ doesn't have integral solutions in $a$ and $b$. 
However, by solving $$\left(\frac{a + b\sqrt{-19}}{2}\right)^3 = x + \sqrt{-19}$$ I got the equation $b(3a^2 - 19b^2) = 8$ which gave me the desired solutions to $E$, namely $(x, y) = (\pm 18, 7)$.
How could I have known which generator for $I$ I should have chosen to get to the right (and only) solutions?
My confusion is in the fact that $$\mathbb{Z}\left[\frac{1 + \sqrt{-19}}{2}\right] = \mathbb{Z} \oplus \mathbb{Z}\frac{1 + \sqrt{-19}}{2},$$ so I thought that my first try would simply do the trick, but it didn't.

To solve $$\left(a+b\frac{1+\sqrt{-19}}{2}\right)^3=x+\sqrt{-19}$$ for integers $a,b$ you should express the right-hand side in terms of the integral basis $1, \frac{1+\sqrt{-19}}{2}$ as well: $$x+\sqrt{-19} = (x-1)+2\frac{1+\sqrt{-19}}{2}.$$
(You could also equate coefficients w.r.t the $\mathbb{Q}$-basis $1,\sqrt{-19}$, but then you equate half-integers.)
(To work efficiently with $\frac{1+\sqrt{-19}}{2}$, use its minimal polynomial $x^2 - x + 5$.)
Equating the coefficients w.r.t. the $\mathbb{Z}$-basis $1, \frac{1+\sqrt{-19}}{2}$, you get $$a^3 - 15ab^2 - 5b^3 = x-1 \qquad \text{and}\qquad b(3a^2 + 3ab - 4b^2) = \mathbf{2}.$$
The second equation does have solutions, namely $(a,b)=(1,1)$ and $(a,b)=(-2,1)$; this leads to $x = \pm 18$.
