1). Having identity does not factor into the proof. Using noncommutativity would require you to restate everything in terms of right or left ideals.
2) Is an ideal generated by an infinite set has in fact infinitely many generators, or does this just mean that such an ideal is simply generated by infinitely many elements, but not necessarily infinitely many generators? The second thing is more correct. Really you should probably be thinking in terms of "finitely generated" and "not finitely generated" (which is a little different from "infinitely generated".)
3) The proof of the on direction is not correct as we have been discussing in the comments. To get you started for the $\impliedby$ direction:
Assume all ideals are finitely generated and let $I_1\subseteq I_2\subseteq\ldots$ be an ascending chain. Let $I=\bigcup_{i=1}^\infty I_i$ be the union of these ideals (and prove it is an ideal, if you haven't already.
Now $I$ is finitely generated by $\{a_1, a_2,\ldots a_n\}$. The important step in logic takes place right here:
prove that there must be a $j\in \mathbb N$ such that $\{a_1, a_2,\ldots a_n\}\subseteq I_j$. Conclude that $I=I_j$, and that $I_k=I$ for all $k\geq j$.
The idea for $\implies$ is the right one, but it has a technical flaw: you do not establish that the chain is strictly increasing. You can do this inductively. (The other posts mentions a fault with assuming $I$ is countably generated; however, I only see that as a disagreement in notation, and what is written doesn't really say that. You can well-order whatever generating set and write the initial segment of countably many generators like that, IMO.)
Your idea that finitely generated ideals are "small" and non-finitely generated ideals are "large" is misleading. Think about $R = \Bbb{Z}[X_1, X_2, \ldots]$, the ring of polynomials over the integers in countably many variables $X_1, X_2, \ldots$ If you define $J_k = \langle X_1, \ldots, X_k\rangle$, then $J_1 \subseteq J_2 \subseteq \ldots$ is a strictly increasing chain of ideals, so $R$ is not Noetherian. The finitely generated ideal $J = \langle 2 \rangle $ properly contains the non-finitely generated ideal $K = \langle 2X_1, 2X_2, \ldots \rangle$. For any prime $p$, the ideal $L_p$ comprising the polynomials with constant term a multiple of $p$ is maximal non-finitely generated.
Best Answer
You have the right idea but your proof isn't correct as stated. It isn't true that each stabilizing chain ends in a maximal ideal. For example, you could just find a Noetherian ring where $0$ isn't maximal and just repeat $0 \subset 0 \subset 0 \subset \dots$
What you should do is define this process:
Start with $0$. If $0$ is maximal then define $I_i = 0$ for $i \geq 1$. Otherwise, find some proper ideal $I_1$ that contains $0$ and put it in the list.
$0 \subset I_1$
If $I_1$ is maximal let $I_i = I_1$ for $i \geq 2$. Otherwise find some proper ideal $I_2$ that contains $I_1$ and put it in the list.
Inductively we have $0 \subset I_1 \subset I_2 \subset \dots \subset I_n \subset \dots$
By the Noetherian property, this chain stabilizes. The ideal it stabilizes to is maximal, or else by construction we would have chosen an ideal properly containing it to succeed it in the chain.
Also note that by a Zorn's lemma argument, every non zero ring has a maximal ideal. They key here is that for Noetherian rings you don't need Zorn's lemma.
EDIT: I was informed that this argument uses the axiom of dependent choice so I will rewrite it here to make this clear.
The axiom of dependent choice states that for any nonempty set $X$ and any entire binary relation $T$ on $X$, there exists a sequence $(x_n)$ such that for all $n \geq 0$, $x_n T x_{n+1}$. A binary relation $T$ on $X$ is entire if for all $x \in X$, there exists $y \in X$ such that $xTy$.
Let $X$ be the set of all proper ideals of a nonzero noetherian ring $R$. Since $R \neq 0$, $X$ is nonempty. Consider the binary relation "<" of strict inclusion. If $R$ has no maximal ideals, then "<" is entire. So by the axiom of dependent choice, if $R$ has no maximal ideals, then we may choose a sequence $(x_n)$ such that
$x_1 < x_2 < \dots < x_n < \dots$
This contradicts the Noetherian property. Hence $R$ has a maximal ideal.