You're absolutely right. More generally, if you have an element $a$ that generates a finite cyclic group $G$, the group is also generated by another element $a^n$ (in multiplicative notation, $n\cdot a$ in additive notation) iff $n$ is coprime to $|G|$, i.e. $\gcd(n,|G|)=1$. This is because in that case, you can write any exponent $k$ in $a^k$ as $k=rn+s|G|$, and thus
$$a^k=a^{rn+s|G|}=\left(a^n\right)^r+\left(a^s\right)^{|G|}=\left(a^n\right)^r\;.$$
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).
Best Answer
Your examples work nicely. Let $\mathbb{Z}_n$ be the cyclic group of order $n$. The following theorem is useful in looking at this sort of situation (taken from Contemporary Abstract Algebra by Gallian, 5th ed.):
So how can we apply this? Well, clearly $1$ is an element of $\mathbb{Z}_n$ of order $n$. Then (now in additive notation to be consistent with the operation in $\mathbb{Z}_n$) $\langle k\cdot 1 \rangle = \langle k \rangle = \langle \gcd(n,k)\cdot 1 \rangle$, and so if $\gcd(n,k)=1$ then $\langle k \rangle = \langle 1 \rangle = \mathbb{Z}_n$.
This also follows by the second part of the theorem, by noting that $|\langle k \rangle |=|k|$. If $\gcd(n,k)=1$ then $|k|=\frac{n}{1}=n=|\langle k \rangle |$ so $k$ generates $\mathbb{Z}_n$. On the other hand, if $m=\gcd(n,k)\ne 1$ then $|k|=\frac{n}{m}=|\langle k \rangle |<n$, so $k$ does not generate $\mathbb{Z}_n$ but rather a subgroup of order $\frac{n}{m}$.