Let $F$ be a field and $G = SL_2(F)$. Let us define the following matrices $$X(b) = \begin{pmatrix} 1 & b\\ 0 & 1 \end{pmatrix}, \quad Y(c) = \begin{pmatrix} 1 & 0\\ c & 1 \end{pmatrix},$$ where $b,c\in F$. It a well-known result that these matrices generate $G$ (Lemma 8.1, Ch. XIII in 3rd edition of Lang's Algebra).
After doing some short calculations, I noticed that the set consisting of all matrices $X(b)$ and the matrix $Y(c_0)$ (for some $c_0\in F^\times$) is already a generating set.
Is there some other way to obtain this result?
AYK
PS. These are the calculations that I mentioned above: $$X(-1/x)Y(x)\cdot X(y)\cdot Y(-x)X(1/x) = Y(-x^2y).$$
I think, doing "these short calculations" is the best method to obtain this result. We may even use only two unipotent generators for finite fields, with a few exceptions. A precise statement is given in Dickson's Theorem. For a finite field $F$, and a generator $\lambda$ for $F$ over $\mathbb{F}_p$, $$ L:=\langle \left( \begin{array}{clcr} 1 & 0\\1 &1 \end{array} \right),\left( \begin{array}{clcr} 1 & \lambda\\0 &1 \end{array} \right) \rangle $$ is indeed $SL_2(F)$, except when $|F| =9$, in which case $L\cong SL_2(5)$. For a reference see here.