$G=\langle x,y\ |\ x^{-1}y^2x=y^{-2}, y^{-1}x^2y=x^{-2} \rangle$ is torsion-free. I have to prove that $G=\langle x,y\ |\ x^{-1}y^2x=y^{-2}, y^{-1}x^2y=x^{-2} \rangle$ is torsion-free.
Some things about this group that I understand are first show that $M=\langle (xy)^2,x^2,y^2 \rangle \unlhd G$, which is fine as $N=\langle x^2,y^2 \rangle \unlhd G$ and setting $z=xy$ and noticing $x^{-1}zx\equiv y^{-1}zy\equiv z^{-1}\ \text{mod}\ N $ and raising square power on both sides tells that $M=\langle z^2,x^2,y^2 \rangle \unlhd G$ and also $M$ is abelian and also torsion-free as it is freely generated by three elements.
Now $G/M$ is a Klein 4- group =$\{M,xM,yM,zM\}$ is also fine.
Now I want to check that if I pick any $g\in G=M\cup x M\cup yM\cup zM$ such that $1\neq g$ and $|g|=n$, then we get some contradiction.
If I let $g\in xM$, say $g=xm$, then I cannot seem to work my way out of here.
What else I see is as $g\in xM\in G/M$, definitely, $o(gM)\ |\ 4$, and thus $g$ can have even order only.
 A: I am not sure how you proved that $M$ is free abelian of rank $3$ but I believe it.
In fact $x^{-1}z^2x=y^{-1}z^2y = z^{-2}$.
For $x^{-1}z^2x = yxyx$ and 
$$(yxyx)(xyxy) = yxyx^2yxy = yxy^2x^{-2}xy = yxy^2x^{-1}y = yy^{-2}y=1,$$
and $y^{-1}z^2y=z^{-2}$ is proved similarly.
So, for $m = x^{2i}y^{2j}z^{2k} \in M$, we have
$(xm)^2 = x^{2+4i}$ which is a nontrivial element of the torsion-free subgroup $M$, so $xm$ is not a torison element. Similarly for $ym$ and $xym$.
A: Let $\Pi$ denote the group admitting $$ \langle x,y \mid x^{-1}y^2x=y^{-2}, y^{-1}x^2y=x^{-2} \rangle$$ as a presentation. Clearly, $$\Pi \simeq \langle x,y,a,b \mid x^{-1}ax=a^{-1} , y^{-1}by=b^{-1}, a=y^2,b=x^2 \rangle.$$ For convenience, we will modify slightly this presentation as $$\Pi'=\langle x,y,a_1, \ldots, a_n,b_1, \ldots, b_n \mid x^{-1}\bar{a} x=\bar{a}^{-1} , y^{-1}\bar{b}y=\bar{b}^{-1}, \bar{a}=y^2,\bar{b}=x^2 \rangle$$ where $\bar{a}$ and $\bar{b}$ denote respectively $a_1 \cdots a_n$ and $b_1 \cdots b_n$. Using Tietze transformations, it is not difficult to notice that $\Pi' \simeq \Pi \ast \mathbb{F}_{n-1} \ast \mathbb{F}_{n-1}$ where the two free groups correspond to $\langle a_2, \ldots, a_n \rangle$ and $\langle b_2, \ldots, b_n \rangle$ respectively. In particular, we deduce that $\Pi$ is torsion-free if and only if so is $\Pi'$. On the other hand, $\Pi'$ can be decomposed as
$$\left( \underset{:=\Pi_a}{\underbrace{\langle x,a_1,\ldots,a_n \mid x^{-1} \bar{a} x=\bar{a}^{-1} \rangle}} \ast \underset{:=\Pi_b}{\underbrace{ \langle y,b_1, \ldots, b_n \mid y^{-1} \bar{b} y = \bar{b}^{-1} \rangle}} \right) / \langle \langle \bar{a}=y^2,\bar{b}=x^2 \rangle \rangle.$$
As above, using Tietze transformations, we notice that $\Pi_a \simeq \Pi_b \simeq BS(1,-1) \ast \mathbb{F}_{n-1}$ where $BS(1,-1)$ is the Baumslag-Solitar group with presentation $\langle p,q \mid pqp^{-1}=q^{-1} \rangle$. In particular, $\Pi_a$ and $\Pi_b$ are torsion-free since so is $BS(1,-1)$: it is an HNN extension over a torsion-free group, namely $\mathbb{Z} \underset{\mathbb{Z}}{\ast}$. (We also have the isomorphism $BS(-1,1) \simeq \langle x,y \mid x^2=y^2 \rangle$, so that we may apply the arguments of this previous question.)
Now, according to Theorem 10.1 in Lyndon and Schupp's book Combinatorial group theory, if a family $R \subset F_1 \ast F_2$ satisfies the small cancellation property $C'(1/8)$, then the quotient $(F_1 \ast F_2) / \langle \langle R \rangle \rangle$ is torsion-free provided that $F_1$ and $F_2$ are torsion-free and $R$ does not contain a cyclic permutation of a proper power.
In our case, $R= \{ \bar{a}y^{-1}, \bar{b}x^{-1} \rangle$ satisfies $C'(1/n)$, so that we deduce from Lyndon and Schupp's theorem that $\Pi'$ is torsion-free whenever $n \geq 8$. A fortiori, $\Pi$ is torsion-free.
For more information on small cancellations for free products, see Section V.9 of Lyndon and Schupp's book.
