I am given the following group: $$G = \langle x_1,x_2,x_3 | x_1^2 = x_2^2 = x_3^2 = e, \langle x_1, x_2 \rangle = \langle x_2, x_3 \rangle = e \rangle,$$ where $\langle a,b \rangle = e$ is the triple relation, meaning that $$aba = bab.$$
I want to prove that the element $x_1 x_3 x_1 x_2$ has infinite order. This seems to me rather trivial, since (intuitively speaking) the triple relations in $G$ cannot reduce the number of generators in $(x_1 x_3 x_1 x_2)^n$.
However, when trying to prove this formally, I could not finish the proof. I tried using induction, playing with the evenness of the number of the generators $x_1, x_2, x_3$ that appear in any power of this element, but without success.