= Solution
We prove the needed <Hilbert basis theorem> step rather than assuming it. Suppose $A$ is a commutative <Noetherian ring> and $I\subseteq A[X]$ is an ideal. The set $J$ of leading coefficients of elements of $I$, together with zero, is an ideal of $A$: align degrees with powers of $X$ before addition, and use scalar multiplication. Choose finitely many generators of $J$, represented by polynomials $f_1,\ldots,f_s\in I$, and let $d$ be the maximum of their degrees. If $J=0$ then $I=0$ and there is nothing to prove.
For each $0\le j<d$, let $J_j$ consist of the coefficients of $X^j$ of elements of $I$ of degree at most $j$. This is also an ideal of $A$; choose finitely many generating coefficients and representing polynomials $g_{j,l}\in I$ of degree at most $j$. The finite collection of all $f_i$ and $g_{j,l}$ generates $I$: for any $h\in I$ of degree $m\ge d$, express its leading coefficient using those of $f_i$ and subtract the matching combination of $X^{m-\deg f_i}f_i$ to lower degree. When the degree becomes $j<d$, use the representatives of $J_j$ to lower it again. Induction on degree eventually leaves zero.
A field is Noetherian because its only ideals are zero and the whole field. Repeatedly applying the proved step gives \b[every ideal of $\boxed{F[X_1,\ldots,X_n]}$ finitely generated.]
The coefficient restriction in the second ring says precisely
$$
R=F+XYF[X,Y].
$$
It contains one and is closed under addition, additive inverses and multiplication: $(a+XYp)(b+XYq)=ab+XY(aq+bp+XYpq)$. To show it is not Noetherian, consider the ideals in $R$
$$
I_d=(XY,XY^2,\ldots,XY^d)_R,\qquad d\ge1.
$$
They form an ascending chain. In any $\sum_{j=1}^d r_jXY^j$ with $r_j\in R$, the coefficient of $X^1$ is a polynomial in $Y$ of degree at most $d$, since the nonconstant part of $r_j$ introduces at least $X^2$. It cannot equal $XY^{d+1}$; hence $I_d\subsetneq I_{d+1}$. If the union ideal were finitely generated, all its generators would lie in one $I_d$, forcing the chain to stabilize. This contradiction proves $\boxed{R\text{ is not Noetherian}}$, the <constant-plus-ideal non-Noetherian subring> example.
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