How to prove that $k[x, xy, xy^2, \dotsc]$ is not noetherian?

Preliminary remark
Given a polynomial $f(x,y)\in k[x,y]$, write $[f]_{i,j }$ for its term $ax^iy^j$ in the monomial $x^iy^j$.
Then for a finite sum $f=\sum_lf_l$ of polynomials we have $[f]_{i,j }=\sum [f_l]_{i,j}$

Proof that $xy^n \notin (x,xy,xy^2,\ldots,xy^{n-1})$
Suppose that $xy^n=\sum _{l=0}^{n-1}g_lxy^l$ with $g_l\in R$.
From the preliminary remark we get $$ xy^n=[xy^n]_{1,n}=[\sum _{l=0}^{n-1}g_lxy^l]_{1,n}\stackrel{prel.rem.}{=} \sum _{l=0}^{n-1}[g_lxy^l]_{1,n}=\sum _{l=0}^{n-1}[g_l]_{0,n-l}xy^l$$ But all $[g_l]_{0,n-l}=0$ since $n-l\gt0$ and all non-constant terms of a polynomial in $R$ involve a positive power of $x$.
Contradiction.

Define $I:=(x, xy, ..., xy^{n-1})$. Any monomial in $I$ must contain one of the generators of $I$ as a factor. Say $xy^n \in I$. Write $xy^n=pxy^i$, where $0 \le i<n$ and $p \in R$. Since $R$ is a domain, we may cancel $xy^i$ from both sides, so that we have $y^{n-i} =p \in R$.

Now, take a general polynomial $f \in R$. We may write is as $c_0 + c_1x + c_2xy +\cdots$, where the $c_i$ are in $k$, and all but finitely many of them are $0$. If $f(x, 0)=0$, then $c_0=c_1=0$, which means either $f$ contains $x$ as a factor or $f=0$. Thus, $y^{n-1} \notin R$.