An Infinite Cyclic Group has Exactly Two Generators: Is My Proof Correct?

You can simplify it greatly, and also do part (2) without contradiction (i.e. directly).

(1) If $G = \langle a \rangle$, then $G$ also equals $\langle a^{-1} \rangle$, because every element $a^n$ of $\langle a \rangle$ is also equal to $(a^{-1})^{-n}$.

(2) If $G = \langle a \rangle = \langle b \rangle$, then $b = a^n$ for some integer $n$, and $a = b^m$ for some integer $m$. These imply that $a = b^m = (a^n)^m = a^{nm}$; since $G$ is infinite cyclic, $nm$ must equal $1$. As an equation in the integers, $nm = 1$ has only two solutions, $(1,1)$ and $(-1,-1)$. These give $b = a$ or $b = a^{-1}$.