3.11 (1969)

Yes, there does (A. P. Goryushkin, Siberian Math. J., 14 (1973), 146–148). Another example is $G = \langle a, b \mid (b^2)^a = b^{-2}, (a^2)^b = a^{-2} \rangle$. Then $N = \langle a^2, b^2, (ab)^2 \rangle$ is an abelian normal subgroup of $G$ and $G/N$ is non-cyclic of order 4. If $a^{2l} b^{2m} (ab)^{2n} = 1$, then after conjugating by $a$ we get $a^{2l} b^{-2m} (ab)^{-2n} = 1$, whence $a^{4l} = 1$. In view of the obvious homomorphism $G \to \mathbb{Z} \wr (\text{Aut}\,\mathbb{Z})$ that maps $a$ to the number $1 \in \mathbb{Z}$ we have $l = 0$. Similarly, $m = n = 0$. Hence $N$ is free abelian of rank 3. The squares of elements outside of $N$ are non-trivial; for example, $(a a^{2l} b^{2m} (ab)^{2n})^2 = a^2 a^{-1} (a^{2l} b^{2m} (ab)^{2n}) a a^{2l} b^{2m} (ab)^{2n} = a^{4l+2} \neq 1$. Hence $G$ is torsion-free. Since the square of any element $x \in G$ belongs to $N$, we have $x^2 (x^2)^a (x^2)^b (x^2)^{ab} = 1$. (V. A. Churkin, 1973.)