Suppose $p$ is prime and the field is $F=\mathbb{Z}_p$ and our group is $G=GL(2,F)$. We know that $|G|=(p^2-1)(p^2-p)$ and I need to find a $2\times2$ matrix in $G$ with order $p^2-1$.
We know that $\mathbb{Z}_p / {0}$ is cyclic so there is a $b$ with $\operatorname{ord}(b)=p-1$. I have good reasons to believe that this matrix $$A=\begin{bmatrix} b & 1 \\ 1 & 0 \end{bmatrix}$$ has indeed order $p^2-1$, but I have no idea how to prove this.
I have found a recursive formula for $A^n$ but that did not help a lot. It would also suffice if I could show in some way that there is a cyclic subgroup of $G$ with order $p^2-1$ but I think finding the generator would be the best way to go. Finding eigenvalues in order to diagonalize this matrix does not quite work either. Any idea would be appreciated.
The matrix you have written cannot have order $p^2-1$ for $p>3$.
I don't believe that it is easy to write down a general expression for such a matrix involving only a primitive element of ${\mathbb F}_p$. A matrix of order $p^2-1$ in ${\rm GL}(2,p)$ has eigenvalues $w$ and $w^p$ over ${\mathbb F}_{p^2}$,where $w$ is a primitive element of ${\mathbb F}_{p^2}$, so its minimal polynomial is $x^2 - (w+w^p)x + w^{p+1} = x^2 - {\rm Tr}(w)x + {\rm N}(w)$, and the companion matrix of that polynomial lies in ${\rm GL}(2,p)$ and has order $p^2-1$.
Note that the determinant of such a matrix is $w^{p+1}$, which is a primitive root of ${\mathbb F}_p$, whereas your proposed matrix has determinant $-1$.