$\mathbb{Z}\rtimes\mathbb{Z}_2$ is generated by an element of infinite order $a$ and an element of order two $b$ satisfying the relation $bab^{-1}=a^{-1}$. Try to find two elements of $\langle x,y\mid x^2,y^2\rangle$ with the same properties which also generate the whole group. Then you can establish an isomorphism.
Edit: I didn't realize you were talking about $x$ and $y$ being in the automorphism group, and not the group itself. I've edited below to correct this.
We can take $\alpha$ as a generator of $C_7$, and then $x(\alpha)=\alpha^5$. So $x^2(\alpha) = \alpha^4$ and $x^{-2}(\alpha) = \alpha^2$.
We can take $\beta$ as a generator of $C_{13}$, and then $y(\beta)=\beta^2$, and so $y^4(\beta) = \beta^3$.
Consider your groups as $G=C_{91}\rtimes C_3$ and $H=C_{91}\rtimes C_3$. We can write $z$ as the generator of $C_{91}$, and $g$ and $h$ as the generator of the $C_3$ subgroups.
Then in $G$, we have $g^{-1}zg=z^{-10}$, since $-10$ is $4\pmod{7}$ and $3\pmod{13}$.
In $H$, we have $h^{-1}zh=z^{16}$, since $16$ is $2\pmod{7}$ and $3\pmod{13}$.
If $f:G\rightarrow H$ was an isomorphism, then we would have $f(z)=z^k$ for some integer $k$. We would also have $f(g)$ is some conjugate of $h$ or $h^2$, which thus acts the same as $h$ or $h^2$ on $z$.
So then
\begin{align*}
z^{-10k} &= f(z^{-10})\\
&= f(g^{-1}zg)\\
&= f(g^{-1})z^kf(g)\\
&= z^{16k}\text{ or }z^{74k}
\end{align*}
This implies $z^{26k}=1$ or $z^{84k}=1$, so that $7$ or $13$ divides $k$ respectively, which means $z^k$ is not a generator of $C_{91}$ in $H$. This is the contradiction.
This same idea can be used whenever you have a semidirect product $A\rtimes B$, and $A$ is abelian and $B$ is cyclic. It can be used to differentiate different semidirect products coming from non-conjugate images of $B$ in $\textrm{Aut}(A)$.
Best Answer
Depends on what you're using as your definition of $D_8$.
One way would be to show that if you take $V=C_2\times C_2$, $\langle \alpha \rangle$ injects into the automorphism group of $V$, which is isomorphic to $GL_2(\mathbb{F}_2)$, by mapping $\alpha$ to $$\left(\begin{array}{cc}1&1\\0&1\end{array}\right).$$ Then show that the semidirect product defined by this automorphism is isomorphic to $D_8$ by whatever presentation you're using.
You could try the same idea using permutation group notation, taking $V=\langle(1,2),(3,4)\rangle$ and mapping $\alpha$ to the automorphism $$\varphi_\alpha:\left\{\begin{array}{l}(1,2)\mapsto (1,2)(3,4) \\ (3,4)\mapsto (3,4)\end{array}\right. .$$
Or yet another variation would be the presentation $$\langle r,s,t|r^2=s^2=t^2=[r,s]=[s,t]=1,r^t=rs\rangle.$$
It all depends how you want to do it, but this automorphism is the key.