As you noted, $N(8) = 64$. Also note that $N(1 \pm \sqrt{-7}) = 8$ and $N(2) = 4$. Since the ring of algebraic integers of $\textbf Q(\sqrt{-7})$ is said to be a unique factorization domain and, as you noticed, $(1 - \sqrt{-7})(1 + \sqrt{-7}) = 2^3 = 8$, this must mean that the two apparently distinct factorizations are in fact incomplete factorizations, just as $4^3$ would be an incomplete factorization of $64$ in $\textbf Z$.
As it turns out, $$\frac{1 + \sqrt{-7}}{2}$$ is an algebraic integer and it belongs in this domain, since its minimal polynomial is $x^2 - x + 2$. Therefore, as numbers, we have $$\left( \frac{1}{2} - \frac{\sqrt{-7}}{2} \right)^3 \left( \frac{1}{2} + \frac{\sqrt{-7}}{2} \right)^3 = 8.$$
However, for ideals, we need to verify that $$\left\langle \frac{1}{2} - \frac{\sqrt{-7}}{2} \right\rangle \not \subseteq \left\langle \frac{1}{2} + \frac{\sqrt{-7}}{2} \right\rangle$$ nor vice-versa. A couple of divisions will quickly confirm that $\langle 2 \rangle$ is a splitting, not ramifying ideal. Therefore, $$\langle 8 \rangle = \left\langle \frac{1}{2} - \frac{\sqrt{-7}}{2} \right\rangle^3 \left\langle \frac{1}{2} + \frac{\sqrt{-7}}{2} \right\rangle^3.$$