Subring of $\mathbb{Z}[x,y]$ generated by $\{yx^i | i\geq 1\}$ is not Noetherian

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Let $R$ be the subring of $\mathbb{Z}[x,y]$ generated by $\{yx, yx^2, yx^3, ...\}$. I proved that $R$ is not Noetherian as follows:

The ideal $(yx^2,yx^3, ...)$ is not finitely generated since any set of generators for this ideal must contain $yx^i$ for each $i\geq 2$. Is this correct?

If it is correct, by the same reasoning, can I conclude that the $R$ can not be finitely generated?

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Since $I$ is generated by monomials, if it were finitely generated then it would have a finite monomial basis. To see this, observe that if $f \in I$ then each monomial of $f$ must also be in $I$. Given a finite monomial basis for $I$, it is easy to see why there is a contradiction.

It should also be the case, by the same reasoning, that $R$ is not finitely-generated. I know that when rings are required to contain $1$, then if $B$ is a finitely generated $A$ algebra and $A$ is Noetherian, then so is $B$ by Hilbert's basis theorem.