Non-Noetherian subring of $\mathbb{C}[X,Y]$

Question

I'm trying to understand the proof to the following problem.

Give an example of a non-Noetherian subring of $\mathbb{C}[X, Y] .$

Proof: Consider the subring $R$ of $\mathbb{C}[X, Y]$ generated (over $\mathbb{C})$ by $X, X Y, X Y^{2}, X Y^{3}, \ldots$. This has a maximal ideal $\mathfrak{m}$ given by the kernel of the ring homomorphism $\phi : R \rightarrow \mathbb{C}$ taking $X$ and $Y$ to zero. Note that this ideal is generated by the monomials $X$ $X Y, X Y^{2},$ etc. It can be shown that if $f$ is in this ideal, then every monomial of $f$ is in this ideal. Thus if this ideal is finitely generated, it is generated by a finite set of monomials (namely, all the monomials appearing in the given finite set of generators). So suppose $\mathfrak{m}$ is generated by a finite set of monomials $c_{i} X^{a_{i}} Y^{b_{i}},$ and let $b$ be larger than all of the $b_{i} .$ Then $X Y^{b}$ cannot lie in $\mathfrak{m},$ which is a contradiction.

I had some questions. Partial answers (answering only a few of the questions) are very welcome! (I will upvote them too)

1. Firstly, we know from the first isomorphism theorem that $R / \text{Ker}(\phi) \cong \text{Im}(\phi) \subset \mathbb{C}$. But how can I know that $\text{Im}(\phi)$ is a field to conclude $\mathfrak{m}$ is maximal? Also, isn't $\text{Im}(\phi) = 0$? $R$ is generated by $X, X Y, X Y^{2}, X Y^{3}, \ldots$, and $\phi(X\times Y^n) = \phi(X)\times \phi(Y)^n = 0$ as $\phi$ sends $X,Y$ to $0$. So for any element of $R$, $\phi$ sends it to $0$?

2. I don't understand how we are using that $\mathfrak{m}$ is maximal.

3. Am I right to think that the reason why $X Y^{b}$ cannot lie in $\mathfrak{m}$ is because $R$ is generated by $X, X Y, X Y^{2}, X Y^{3}, \ldots$, thus for a higher power of $Y$ than $b_i$ to be in $\mathfrak{m}$, it must be divisible by $X^2$?

4. How can we show that "if $f$ is in this ideal, then every monomial of $f$ is in this ideal"?

Thanks in advance...