Alternate approach via the induced automorphism.
The relation $xyx^{-1}=y^2$ shows that $x$ normalizes the cyclic subgroup $\langle y\rangle$, and conjugation by $x$ acts on $\langle y\rangle$ as the map $\varphi(y)=y^2$, an automorphism given by squaring. Conjugation by $x^5$ is the same as applying $\varphi$ five times, that is $\varphi^5$. Since $x^5=e$, conjugation by $x^5$ is the identity map, so $\varphi^5=\text{id}$ on $\langle y\rangle$. This gives
\[\varphi^5(y)=y^{2^5}=y^{32}=y \implies y^{31}=e,\]
so the order $n$ of $y$ divides $31$. As $31$ is prime and $y\ne e$ rules out $n=1$, we get $n=31$.
As a check, if $n=31$ the automorphism $\varphi(y)=y^2$ on the cyclic group of order 31 has order equal to the multiplicative order of $2$ modulo $31$. Computing: $2^1=2$, $2^2=4$, $2^3=8$, $2^4=16$, $2^5=32\equiv 1\pmod{31}$. So $2$ has multiplicative order exactly $5$ modulo $31$, matching $x^5=e$ being the relation that forces $\varphi^5=\text{id}$, with no smaller power of $x$ forcing $y=e$. This confirms $y$ genuinely has order 31.
\[\boxed{\text{Order of } y = 31}\]