Prove that $G = \langle x,y\ |\ x^2=y^2 \rangle $ is torsion-free. 
Prove that $G = \langle x,y\ |\ x^2=y^2 \rangle $ is torsion-free.

Here $x^2$ is central as $x^2y=yx^2$ similarly $y^2$ is central. Apart from this  I do not know how to proceed. Taking any arbitrary word in $G$ does not help. 

As noticed here, $G= \langle x,y \mid x^2=y^2 \rangle$ is the fundamental group of the Klein bottle $K$. 
Argument 1: Any subgroup $H \leq G$ comes from a cover $C \to K$ where $\pi_1(C) \simeq H$. But $C$ is a surface, and, by looking at the abelianizations of the closed surface groups, we deduce that $C$ cannot be compact. On the other hand, it is known that the fundamental group of any non-compact surface is free. Consequently, $\pi_1(C) \simeq H$ cannot be a finite cyclic group. For more information, see here.
Argument 2: The universal cover of $K$ is the plane $\mathbb{R}^2$, so that $K$ is aspherical: $K$ is a classifying space of $G$. Because $K$ is finite-dimensional, $G$ must be torsion-free.
Argument 3: As noticed here, using the fact that there exists a two-sheeted covering $\mathbb{T}^2 \to K$, we deduce that $\langle x^2, xy^{-1} \rangle$ is a subgroup of index two in $K$ isomorphic to $\mathbb{Z}^2$. Since $$G / \langle x^2,xy^{-1} \rangle = \langle y \mid y^2=1 \rangle,$$ it is sufficient to prove that $y$ has infinite order in $G$. As Derek Holt did, one may consider the quotient $G/ \langle \langle xy^{-1} \rangle \rangle$, which is infinite cyclic.
Argument 4: $G$ is an amalgamated product $\mathbb{Z} \underset{\mathbb{Z}}{\ast} \mathbb{Z}$ of torsion-free groups, so it is torsion-free.
