Basic proof that $GL(2, \mathbb{Z})$ is not nilpotent I need to show that $GL(2,\mathbb{Z})=\left\{ \begin{pmatrix} a&b \\ c&d \end{pmatrix} \mid a,b,c,d \in \mathbb{Z}, ad-bc = \pm 1\right\}$ is not nilpotent. I have seen this question but the answer given there is too advanced for where I am currently in my studies.
In order to show something is not nilpotent, all I have at my disposal is to show that the upper central series does not terminate or that it is not solvable (using a derived series that does not terminate or a subnormal series that either does not terminate or whose factors are not all abelian.)
This seems a little messy, and I've never used a direct approach to show that something was not nilpotent.
What is the best approach to this proof without appealing to $p$-groups, Sylow Groups, Galois Theory, or free abelian groups?
Thank you.

One possibility is to show that there is a proper, self-normalizing subgroup,
$$
\left\{ \begin{pmatrix} a&b \\ 0&d \end{pmatrix} : a, d = \pm 1, b \in \mathbb{Z}\right\}
$$
being perhaps the easiest candidate.
