If there exists $m,n\geq2$ such that $x^{m}y^{n}=yx$, for any $x,y\in M$, then prove that the operation is commutative 
Let $M$ be a set, not null, and $*$ an operation that is associative. If there exist $m,n\geq2$ such that $x^{m}y^{n}=yx$, for any $x,y\in M$, then prove that the operation is commutative. 

Now, I obtained a result by $x{\rightarrow}yxy^{-1}$ such that $yx=(yxy^{-1})^my^n=yx^my^{n-1}=yx^my^ny^{-1}=yyxy^{-1}$ and from here $yx=yyxy^{-1}{\Rightarrow}xy=yx$.

However, this works only if my set is a group. How can I adapt this to the form of the set?

 A: In the four Lemmas that follow, $z$ is an arbitrary element of $M$. 

Lemma $1$: $z^2=z^{m+n}$.

Proof: $z^2=zz=z^{m}z^{n}=z^{m+n}.$ 

Lemma $2$: $z^3=z^{2m+n}.$

Proof: $z^3=zz^2=z^{2m}z^{n}=z^{2m+n}.$

Lemma $3$: $z^2=z^2\cdot z^{m-1}$.

Proof:
\begin{align}
z^2
  &\stackrel1=z^{m+n}
\\&=z^3z^{m+n-3}
\\&\stackrel2=z^{2m+n}z^{m+n-3}
\\&=z^{m+n}z^{2m+n-3}
\\&\stackrel1=z^{2}z^{2m+n-3}
\\&=z^{m+n}z^{m-1}
\\&\stackrel1=z^{2}z^{m-1}.
\end{align}

Lemma $4$: $
z^2=z^2\cdot z^{n-1}.
$

Proof: Should be obvious by the symmetry of the assumptions in $m,n$.

With these Lemmas under our belt, we can solve the problem at hand.
\begin{align}
xy
  &=y^mx^n
\\&=x^{mn}y^{mn}
\\&=x^{m-2}x^{2}(x^{n-1})^my^{n-2}y^2(y^{m-1})^n
\\&\stackrel{3}=x^{m-2}x^{2}(x^{n-1})^my^{n-2}y^2
\\&\stackrel{4}=x^{m-2}x^{2}y^{n-2}y^2
\\&=x^my^n
\\&=yx
\end{align}
In step $\stackrel{3}=$, we apply Lemma $3$ a total of $n$ times with $z=y$, each time eliminating one copy of $y^{m-1}$ next to the $y^2$. In $\stackrel{4}=$, we apply Lemma $4$ $m$ times with $z=x$. 
Note $x^0$ refers to the empty product, so $x^{m-2}$ is valid (here is where we need $m,n\ge 2$). This is not an element of $M$, but the absence of a product.
