Using a mixture of computation and thought I believe that I have established that this group is indeed torsion-free. I don't know of any general approach to solving that particular problem. Even if the group is hyperbolic (which this example is not, because it has free abelian subgroups of rank $2$), I am not aware of any practical implementable algorithm for deciding torsion-freeness.

Here is some Magma code for this problem. Following Pace Nielsen's suggestion that there might be an easily identifiable subgroup of index $32$, I found such a subgroup $K$ of index $4$, and we can compute a presentation of it on its four defining generators.

    > G<x,y,z>:=Group<x,y,z|x*y^-1*x^-1=z^2*y, x*z^-2=z^2*x, x*y*x^-1*y=z^2,
    >                       y^2*x*z=z*x >;
    > K<a,b,c,d> := sub<G | x^2, z^2, x*z*y^-1, y^2>;
    > Index(G,K);
    4
    > Rewrite(G,~K);
    > K;
    Finitely presented group K on 4 generators
    Index in group G is 4 = 2^2
    Generators as words in group G
        a = x^2
        b = z^2
        c = x * z * y^-1
        d = y^2
    Relations
        (c^-1, a) = Id(K)
        (a^-1, b) = Id(K)
        (a^-1, d^-1) = Id(K)
        (d^-1, b^-1) = Id(K)
        (b, c) = Id(K)
        d * c * b^-1 * d^-1 * c^-1 * b^-1 = Id(K)
        b^-1 * a * c^-1 * a^-1 * b * c = Id(K)

We see that all pairs of generators of $K$ commute except for $c$ and $d$. The final relation collapses completely, and the preceding one is equivalent to $dcd^{-1}c^{-1} =b^2$, so $K$ is a direct product of $\langle a \rangle$ and a torsion-free nilpotent group of class $2$ generated by $b,c,d$. So $K$ is torsion-free.

    > Transversal(G,K);
    {@ Id(G), x, y, z @}

So a torsion-element would have to have order $2$ and be of the form $xk$, $yk$ or $zk$ for some $k \in K$. I spent some time trying to prove that this does not happen, but eventually found an easy way to do it by calculating in a finite quotient. We let $K_2$ be the kernel of the map of $K$ on the to the largest elementary abelian quotient of $K$, which has order $16$, and then check for torsion elements in $G/K_2$. We see that there are none, so $G$ is torsion-free.

    > PK, phi := ElementaryAbelianQuotient(K,2);
    > Order(PK);
    16
    > K2 := Kernel(phi);
    > Index(K,K2);
    16
    > T2 := Transversal(K,K2);
    > exists{k : k in T2 | (x*k)^2 in K2 };
    false
    > exists{k : k in T2 | (y*k)^2 in K2 };
    false
    > exists{k : k in T2 | (z*k)^2 in K2 };
    false