I think you are getting confused along the way.
You want to show that if $G$ is an infinite cyclic group, then it has exactly two generators. This can be done by showing two things: that there are at most two generators, and then exhibiting two generators.
Exhibiting two generators is easy: if $G=\langle a\rangle$, then $a$ and $a^{-1}$ both generate; and $a\neq a^{-1}$, since $a$ has infinite order.
The bulk of your argument is an attempt at showing the other direction, namely you are trying to show:
If $a^n$ generates $G$, then $n=1$ or $n=-1$.
You analyse this correctly until the end of the paragraph that begins with a parenthetical remark. You successfully conclude $n=1$ or $n=-1$. So you are done.
But then you seem to be getting confused, and continue to argue; you are already done showing that there are at most two generators, so that's where the proof should end.
If the second part of the proof was meant to be what my first part was, then you are not clear in the first part. There should be an explicit statement where you say that your argument shows there are at most two generators. Finally, there is also the issue of noting that $a\neq a^{-1}$ (which is easy, but needs to be said).
A group $G$ typically have many different generating sets, that is subsets $S\subseteq G$ with $\langle S\rangle = G$, and among these are sets of different cardinality (for example, trivially $S=G$ is a generating set).
Therefore it makes little sense to speak of the number of generators of a group (or even of the set of generators).
We do speak of the free group $F_S$ generated by the set $S$ and for such a free group, the set $S$ is a canonical choice of generators. Nevertheless, the free group over $S=\{a,b\}$ is also generated by $\{a,b,1,a^{-2},bab\}$.
In general, when we speak of a group generated by $n$ elements, we mean a group $G$ that allows an epimorphism $f\colon F_S\to G$ where $|S|=n$, typically given by a presentation $G=\langle a_1,\ldots,a_n\mid\ldots\rangle$. Depending on the author, it may additionally be understood that $f|_S$ is injective, that is that the generators of $G$ obtained this may are actually an $n$-element set. But I guess it is often not the case that this distinction is made.
In summary, "a group generated by $n$ elements" should usually be more precisely called "a group that has at least one generating set of at most $n$ elements".
For example, a cyclic group is a group generated by a single element. We do however also count the trivial group as cyclic, even though it can in fact be generated by zero elements.
Best Answer
The answer to all your questions is yes. By definition a cyclic group is a group which is generated by a single element (or equivalently, by a subset containing only one element). Such an element is called a generator.
$(\mathbf{Z},+)$ of course has infinitely many generating subsets, be it only because any subset containing $1$ or $-1$ is generating, and there are of course infinitely many such subsets. There are more interesting generating subsets however, such as those containing two relatively prime integers.