Here is a further reference, extending the result you have linked.
**Theorem:** If $k$ is square-free and not equal to $1$, the elliptic curve $y^2 = x^3+k$ has no rational torsion points.
_Proof:_ See A. Knapp, Elliptic Curves, Princeton Univ. Press, 1992, Theorem $5.3$.
More generally, there is indeed a classification result as follows.
**Theorem:** Let $k=m^6 \cdot k_0$, where $m, k_0\in \Bbb Z$ and $k_0$ is free of sixth power prime factors. Then the torsion subgroup of $E\colon y^2=x^3+k$ over $\Bbb Q$ is given as follows: $$ E_{\rm tors}(\Bbb Q)= \begin{cases} \Bbb Z/6\Bbb Z & \text{ if } k_0= 1, \\\ \Bbb Z/3\Bbb Z & \text{ if $k_0$ is a square dierent from $1$}, \\\ \Bbb Z/3\Bbb Z & \text{ if $k_0=−432$}, \\\ \Bbb Z/2\Bbb Z & \text{ if $k_0$ is a cube dierent from $1$}, \\\ \\{\mathcal O\\} & \text{ otherwise.} \end{cases} $$