I want to prove the following fact:
Let $G$ be a finite group and let $x, y\in G$ such that $[x, y] \in Z(G)$. Then $[x, y^s] = [x, y]^s$ for every $s\in \mathbb{Z}$. If we assume that $[x^r, y^s] \in Z(G)$ for every $r, s\in \mathbb{Z}$ we have $[x^r, y^s] = [x, y]^{rs}$.
But I am not sure my proof is correct. And even if it may be correct - is there a nicer way to prove it? Here is my attempt:
Proof: We will first show $[x, y^s] = [x, y]^s$ by induction on $s$ and using commutator formulas. For $s=1$ we obviously have $[x, y^1] = [x, y]^1$. So suppose that $[x, y^s] = [x, y]^s$ for some fixed $s$. Then $$[x, y^{s+1}] = [x, yy^s] = [x, y^s][x, y]^y = [x, y]^s[x, y] = [x, y]^{s+1},$$ where we have used our induction hypothesis and the fact that $[x, y] \in Z(G)$.
Now suppose $[x^r, y^s] \in Z(G)$ for every $r,s\in \mathbb{Z}$. We will show $[x^r, y^s] = [x, y]^{rs}$ by induction on $r$. For $r=1$ we have by what we have just shown that $[x, y^s] = [x, y]^{s}$. So suppose that it holds for some fixed $r$ that $[x^r, y^s] = [x, y]^{rs}$. Then (similar to before) $$[x^{r+1}, y^s] = [x^rx, y^s] = [x^r, y^s]^x[x, y^s] = [x^r, y^s][x, y]^{s} = [x, y]^{s+rs} = [x, y]^{s(r+1)},$$ and we are done.
Your proofs are fine, but for one typo: under the conventions $[a,b] = a^{-1}b^{-1}ab, a^c = c^{-1}ac$, the formulas you implicitly use are:
$$[a, bc] = [a,c][a,b]^c, \qquad [ac, b] = [a,b]^c[c,b]$$
Applying the first gives $[x,yy^s] = [x,y^s][x,y]^{y^s}$ (rather than $[x,y^s][x,y]^y$), but since $[x,y] \in Z(G)$, $[x,y]^{y^s} = [x,y]^y = [x,y]$.