Let $g$ be a Lie algebra with enveloping algebra $U(g)$ and $Z(g)$ the center of $U(g)$.
Question 1: When is $Z(g)$ noetherian?
Question 2: When is $U(g)$ a finitely generated $Z(g)$-module? Is this true when $g$ is finite dimensional?
For Question 2, I think the answer is "very rarely" in characteristic 0 (unless $\mathfrak{g}$ is abelian).
For example, let $\mathfrak{g} = \langle x, y, z\rangle_{\mathbb{C}}$ with relation $[x,y] = z$ and $[x,z] = [y,z] = 0$. It follows that $Z(\mathfrak{g}) = \mathbb{C}[z]$. Indeed, let $$ a = \sum_{m, n, l\in \mathbb{N}} c_{m, n, l}x^my^nz^l\in Z(\mathfrak{g}),\quad c_{m,n,l}\in \mathbb{C}. $$ Then we have $$ [x,a] = \sum_{\substack{m, n, l\in \mathbb{N} \\ n \ge 1}}n c_{m, n, l}x^my^{n-1}z^{l+1} $$ so $c_{m,n,l} = 0$ unless $n = 0$. Similarly, $c_{m,n,l} = 0$ unless $m = 0$.
However, in characteristic $p > 0$, for an arbitrary Lie algebra $\mathfrak{g}$ we have $$ U(\mathfrak{g}^p) = \mathbb{k}[x^p\;;\;x\in \mathfrak{g}]\subset Z(\mathfrak{g}) $$ so $U(\mathfrak{g})$ is a finite $Z(\mathfrak{g})$-module provided that $\mathfrak{g}$ is finite dimensional.