I am facing the problem of proving $z^2 \equiv 1\pmod 8$ for all odd $z\in\mathbb{N}$.

What I know is that for each $z \in \mathbb{N}$, $z = m^2-1$, for $m\in\mathbb{N}$.

What I also know is that in a least residue system there are the following solutions for all $z\in\mathbb{N}$.

The solutions are: $$n \equiv 1,3,5,7$$

Which lead to solutions of the initial problem: $$n = 8 \cdot k + 1\\ n = 8 \cdot k + 3 \\ n = 8 \cdot k + 5 \\ n = 8 \cdot k + 7$$ where $k\in\mathbb{N}$.

My question now is how I can transfer my knowledge about the solutions for all $n\in\mathbb{N}$ to the key problem where $n$ must be an odd number and the proofs needs to be done for all odd $n$.

I also know that $8 = 2^3$ which leads to $$(2\;\cdot\;m\;-\;1)^2\;\equiv\;1\;\pmod 8$$ $$(2\;\cdot\;m\;-\;1)^2\;\equiv\;1\;\pmod {2^3}$$ but I do not know what this tells me right now.