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

Let $N$ be the central subgroup $\langle x^2 \rangle$ of $G$. Then $G/N$ is a group with presentation $\langle x,y \mid x^2=y^2=1 \rangle$, which is the infinite dihedral group. It has three conjugacy classes of torsion elements, with representatives $1$, $x$ and $y$, each of which powers into $x^2$, so it is sufficient to prove that $x^2$ is not a torsion element.

One way to see that easily is that $\langle x,y \mid x=y \rangle$ of $G$ is an infinite cyclic group with generator $x$. Since $x$ is not a torsion element in this quotient group of $G$, it cannot be one in $G$ either.

Note $\phi : G \to G/Z$ be the canonical homomorphism where $Z= \langle x^2 \rangle \le G$. Then order of image of $x$ and $y$ is $2$ (because $x^2=y^2$), so $2$ must divide $o(x)$ and $o(y)$,

Now if possible, let $x$ has finite order in $G$ and if that happens, then we may let $o(x)=2a, o(y)=2b$, then this will imply $o(x^2)=o(y^2)=a=b$, i.e. both $x$ and $y$ has equal order say, $n$,

and if that happens then under the group homomorphism $\xi : G \to \langle t \rangle \cong$ $\mathbb{Z}$ such that both $x$ and $y$ goes to $t$ order of image of $x$ , $o(\xi(x))=o(t)$ must divide $n$, but $o(t) = \infty$, contradiction to our assumption, so $n $ must be infinity.

Thus $x$ cannot be a torsion element and by similar reasoning $y$ cannot be a torsion element in $G$, which implies neither $x^2$ nor $y^2$ can be a torsion element in $G$ (as that would imply $x$ or $y$ is torsion resp.). And as suggested by Derek, $G/Z$ has only three conjugacy classes of torsion elements,namely $Z,xZ,yZ$ and $1^2,x^2,y^2 \in Z$ so we are done.

P.S.- This proof is essentially same as Derek's, I just wanted to write it out for my own satisfaction. Thanks! I will be happy if it ll help you in any way. Look more into this group. It has interesting properties.

Here are some links-
G is not free
G is solvable

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.