Let $G$ be a group. We let $G_1:=G$, we let $G_2 := [G,G]$ the commutator subgroup (i.e. the subgroup generated by all the $[a,b]$ such that $a,b\in G$). Inductively one can define $G_k = [G_{k-1},G]$.
We say that $G$ is $k$-step nilpotent if $G_{k+1}$ is trivial.
Let $G$ be a $k$-step nilpotent group. Let $g\in G$ be an element such that $g^n=1$ and let $s\in G$.
Can we conclude that $[s,g]$ is of finite order?
The answer is yes for small $k$'s. If $k=1$, then $G$ is abelian and there's nothing to prove. If $k=2$ this is also true, for simplicity we prove this for $n=2$ (the same proof holds for general $n$)
we have $$[s,g]^2 = sgs^{-1}g^{-1} [s,g]$$
Since $G$ is $2$-step nilpotent, $[s,g]$ commutes with everything, in particular with $g^{-1}$ so we have
$$[s,g]^2 = sgs^{-1} [s,g]g^{-1}=sgs^{-1}sgs^{-1}g^{-1}g^{-1}=sg^{2}s^{-1}g^{-2}=ss^{-1}=1$$
Can we prove this for higher order of nilpotency?
Note that this argument can be generalized, using it one can show that for any $s\in G_{k-1}$, $[s,g]^n=1$.
Best Answer
Modulo correcting the nomenclature, as noted in comments, $$[x,y] = xyx^{-1}y^{-1} = x(yx^{-1}y^{-1})$$ is a product of two conjugates of $x$; if $x$ is torsion, then this is a product of two torsion elements.
In a nilpotent group, the torsion elements form a subgroup. See here, but note that the answer there uses a different commutator convention, namely that $[x,y]=x^{-1}y^{-1}xy$.
Please note the comments on common notation (which clashes with yours) and nomenclature (your definition of $s$-step nilpotent is off).