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.

Because I'm lazy, I'll write $G = \operatorname{GL}(2,\mathbb{Z})$.
From your comments/messages in the chat, I gather that you found the center $Z(G) = \{I,-I\}$ where $I$ is the identity matrix.
Now, consider the upper central series:
$$1 = Z_0 \trianglelefteq Z_1 \trianglelefteq Z_2 \trianglelefteq \cdots \trianglelefteq Z_i \trianglelefteq \cdots$$
where $$Z_{i+1} = \{ x \in G \mid \forall y \in G: [x,y] \in Z_i\}= \{ x \in G \mid \forall y \in G: x^{-1}y^{-1}xy \in Z_i\}.$$ In particular, observe that
$$Z_1 = \{ x \in G \mid \forall y \in G: [x,y] \in Z_0\} = \{ x \in G \mid \forall y \in G: xy = yx \} = Z(G) = \{I,-I\}.$$
Now, onwards to $Z_2$. 
$$Z_2 = \{ x \in G \mid \forall y \in G: [x,y] \in Z_1\} = \{ x \in G \mid \forall y \in G: (xy = yx \text{ or } xy = -yx) \}.$$
Let us examine the second condition: $xy = -yx$. This means that $y = -x^{-1}yx$, which gives 
$$\operatorname{trace}(y) = \operatorname{trace}(-x^{-1}yx) = -\operatorname{trace}(x^{-1}yx) = -\operatorname{trace}(xx^{-1}y) = -\operatorname{trace}(y)$$ hence $\operatorname{trace}(y) = 0$. So $x$ must commute with all matrices $y$ having $\operatorname{trace}(y) \neq 0$. We claim that the only matrices $x$ for which this holds, are $I$ and $-I$ (see below for a proof of this claim).
Thus we end up with $Z_2(G) = Z(G)$ again! So if we repeat these steps again, we'll find that $Z_3 = Z(G)$, and again that $Z_4 = Z(G)$, etc: $Z_i = Z(G) = \{I,-I\}$ for all $i \geq 1$. 
But then the upper central series never terminates; and hence $G$ is not nilpotent!
Proof of claim: Clearly it holds for $x = I$ or $x = -I$, since these groups are in the center of $G$. 
Now, let $x = \begin{pmatrix}x_1 & x_2\\x_3 & x_4\end{pmatrix}$. Then we must have that
$$\begin{pmatrix}x_1 & x_1+x_2\\x_3 & x_3 + x_4\end{pmatrix} = \begin{pmatrix}x_1 & x_2\\x_3 & x_4\end{pmatrix}\begin{pmatrix}1 & 1\\
0 & 1\end{pmatrix} = \begin{pmatrix}1 & 1\\
0 & 1\end{pmatrix}\begin{pmatrix}x_1 & x_2\\x_3 & x_4\end{pmatrix} = \begin{pmatrix}x_1 +x_3& x_2+x_4\\x_3 & x_4\end{pmatrix},$$
hence $x_3 = 0$ and $x_1 = x_4$. Similarly,  commuting $x$ with $\begin{pmatrix}1 & 0\\
1 & 1\end{pmatrix}$ you find that $x_2 = 0$ and $x_1 = x_4$. But since $x \in G$, then the only matrices satisfying this are $I$ and $-I$.
