In Suzuki's Group Theory I, Theorem 6.21 says
Let $p$ be an odd prime number, and let $\lambda$ be an element of $F$ which is algebraic over the prime field $F_0=GF(p)$. Set $E=F_0(\lambda)$. Let $G$ be defined as follows: $$G=\left\langle\left(\begin{matrix}1&0\\1&1\end{matrix}\right),\quad\left(\begin{matrix}1&\lambda\\0&1\end{matrix}\right)\right\rangle$$ Then, either $G$ is isomorphic to $SL(2,E)$, or we have $p=3$, $\lambda^2=-1$, $E=GF(9)$, and $G\cong SL(2,5)$.
The first idea of the proof confuses me. It says
The group $G$ is a subgroup of $SL(2,E)$ generated by two non-commutative elements of order $p$. So, by Theorem 6.17, either $G\cong SL(2,p^n)$, or $p=3$ and $G\cong SL(2,5)$.
In Theorem 6.17, another case for subgroups of $SL(V)$ is that $G$ is congruent to $\langle SL(2,p^n),d_\pi\rangle$, where $d_\pi=\left(\begin{matrix}\pi&0\\0&\pi^{-1}\end{matrix}\right)$. Why is this case not possible for our $G$?
This other group $\langle SL_2(p^n),d_\pi\rangle$ is not defined over $GF(p^n)$, as $\pi$ lies in $GF(p^{2n})$ and not in $GF(p^n)$. Thus $G$ cannot be this group.
Edit: as per the comments, this group is eliminated in the proof because it has a normal subgroup of index $2$. As such, it cannot be generated by two elements of order $p$, as all elements of order $p$ are contained in the normal subgroup.