$\DeclareMathOperator\SL{SL}\DeclareMathOperator\GL{GL}$Let $G_2(\mathbf{Z}_p):=\begin{pmatrix}
1+p\mathbf{Z}_{p}	&  \mathbf{Z}_{p}\\
 p\mathbf{Z}_{p}	& 1+p \mathbf{Z}_{p}
\end{pmatrix}$. Then it is a Sylow pro-$p$ subgroup of $\GL_2(\mathbf{Z}_p$) and $S_2(\mathbf{Z}_p):=G_2(\mathbf{Z}_p) \cap \SL_2(\mathbf{Z}_p)$ is a Sylow pro-$p$ subgroup of $\SL_2(\mathbf{Z}_p$). It's well-known that $S_2(\mathbf{Z}_p)$ is a $2$-generator pro-$p$ group. For exmaple, the set $\{\begin{pmatrix}
	1 & 1 \\
	0 & 1
\end{pmatrix} ,\begin{pmatrix}
1& 0 \\
p& 1
\end{pmatrix} \}$ is a topological generating set. Moreover, in Prop. 3.1.1 of [The image of Galois representations attached to elliptic curves with an isogeny][1] it's proved that if $A,B\in S_2(\mathbf{Z}_p)$ such that the image of $A$ in $\GL_2(\mathbf{F}_p)$ is non-trivial and the image of $B$ in $\GL_2(\mathbf{F}_p)$ is trivial and the image of $B$ in $\GL_2(\mathbf{Z}/p^2\mathbf{Z})$ is not upper triangular, then the set $\{A,B\}$ topologically generates $S_2(\mathbf{Z}_p)$.

**Question:** Can $S_2(\mathbf{Z}_p)$ be generated by two matrices $A,B$ such that the images of $A,B$ in $\GL_2(\mathbf{F}_p)$ are both non-trivial?

As pointed out in the answer, the answer to question is No in a trivial way. What about the case assuming further that **the eigenvalues of $A,B$ are all $1$**?

  [1]: https://muse.jhu.edu/pub/1/article/485931/summary