$|x|=7$,$|y|=3$ then $|\langle x,y \rangle |=21$

Question

Let $|G|=168$. Let $x,y \in G$ with $|x|=7$, $|y|=3$. Show that if $yxy^{-1}=x^2$ then $|\langle x,y \rangle |=21$.

What I have done so far: since $\langle x \rangle \leq \langle x,y \rangle $ and $\langle y \rangle \leq \langle x,y \rangle $ then $7$ and $3$ divide $|\langle x,y \rangle |$, and $21 \leq |\langle x,y \rangle |$. Now, $|\langle x,y \rangle |$ must divide $168$ and is a multiple of $21$ so it is either $21$ or $84$. I am not sure how to use $x^{-1}yx=y^2$ to conclude that it is of order $21$. Any hints appreciated.

Answer 1

Write $yx=x^2y$, so that the group $H=\langle x,y\rangle$ can be written as $$ H=\{x^iy^j\mid 0\le i\le 7, 0\le j\le 2\}. $$ So we have $|H|=21$.