Often what matters is not whether the number ring in question is a UFD but whether its class number is relatively prime to a certain key number. In the case of Fermat's Last Theorem, these factoring methods work well for $x^p + y^p = z^p$ for regular primes, i.e., primes for which $p$ does not divide the class number of the ring of integers of $\mathbb{Q}(\zeta_p)$.
For the Bachet-Mordell equation $y^2 + k = x^3$ (I suppose $k$ is squarefree), the favorable case is $k \equiv 1,2 \pmod 4$ (so that $\mathbb{Z}[\sqrt{-k}]$ is the full ring of integers of $\mathbb{Q}(\sqrt{-k})$) and that the class number of $\mathbb{Q}(\sqrt{-k})$ is prime to $3$. In the particular case $k = 5$, you're in luck: $5 \equiv 1 \pmod 4$ and the class number is $2$, which is as good as $1$!
What exactly do we win when this happens? By some happy coincidence I wrote up some notes on precisely this: please see Theorem 7 here. (Also see the "tables" on page 9.)