There exists no such non-noetherian local ring.
Below I assume by contradiction that we have such a ring.
(a) The first observation is a particular case Proposition 1.2(a) in your reference to Armendariz: for every $r\in R$, we have $r\mathfrak{m}\in\{\mathfrak{m},\{0\}\}$. Indeed, $r\mathfrak{m}$ is a quotient of $\mathfrak{m}$; if it's nonzero, it's quotient by a proper submodule and hence is also not noetherian, which implies $r\mathfrak{m}=\mathfrak{m}$.
(b) case when $R$ is a domain (I just expand your argument). Choose $r\in\mathfrak{m}\smallsetminus\{0\}$. Since $R$ is a domain, by (a) we have $r\mathfrak{m}=\mathfrak{m}$, and in particular so $r\in r\mathfrak{m}$, and since $R$ is a domain this implies $1\in\mathfrak{m}$, a contradiction.
(c) let us check that $R$ is necessarily of Krull dimension 0. Indeed if $P$ is a nonmaximal prime ideal, then $P\neq\mathfrak{m}$ and hence is finitely generated, so $R/P$ is a counterexample to (b).
(d) Now we use Proposition 1.2(b) in Armendariz: $P=\mathrm{Ann}_R(\mathfrak{m})$ is prime (as you've already mentioned). The argument is easy: indeed, for $x,y\notin P$, by (b) we have $x\mathfrak{m}=y\mathfrak{m}=\mathfrak{m}$, which implies $xy\mathfrak{m}\neq 0$, so $xy\notin P$.
(e) Combining (c) and (d), the only option for $\mathrm{Ann}_R(\mathfrak{m})$ is that it's equal to $\mathfrak{m}$. Thus $\mathfrak{m}^2=\{0\}$. Then $\mathfrak{m}$ is an infinite-dimensional vector space over the field $R/\mathfrak{m}$, and all its hyperplanes are ideals. In particular they fail to be finitely generated, and this is a contradiction.
Edit 1: the argument can be extended to show that there is no commutative ring at all with these conditions (non-noetherian such that all non-maximal ideals are finitely generated). That is, assuming $R$ local is unnecessary. In other words, a commutative ring is noetherian if and only if all its non-maximal ideals are finitely generated ideals.
Indeed, (a),(b),(c),(d) work with no change for every given infinitely generated maximal ideal $\mathfrak{m}$. Let us adapt (e):
(e') for every infinitely generated maximal ideal $\mathfrak{m}$, by combining (c) and (d), its annihilator is another maximal ideal $\mathfrak{m}'$. Then $\mathfrak{m}$ can be viewed as a $R/\mathfrak{m}'$-vector space. Hence the lattice of ideals contained in $\mathfrak{m}'$ can be identified to the lattice of $R/\mathfrak{m}'$-vector subspaces of $\mathfrak{m}$. In particular, the condition that all its elements except whole $\mathfrak{m}$ are noetherian, implies that $\mathfrak{m}$ has finite dimension (as vector space over $R/\mathfrak{m}'$), hence is finitely generated as an ideal, a contradiction. We deduce every maximal ideal is finitely generated, and hence (since all non-maximal ones are also finitely generated by assumption) that $R$ is noetherian.
Edit 2:
As mentioned by Keith in the comments, the non-local case can be handled in an even easier way:
let $R$ be a ring (commutativity is unnecessary) in which every non-maximal left ideal is finitely generated, and having at least two maximal left-ideals. If $\mathfrak{m}$ is a maximal left-ideal and $\mathfrak{m}'$ is another one, then $\mathfrak{m}\cap \mathfrak{m}'$ is finitely generated, and $\mathfrak{m}/(\mathfrak{m}\cap \mathfrak{m}')\simeq (\mathfrak{m}+\mathfrak{m}')/\mathfrak{m}'=R/\mathfrak{m}'$ is a simple module, so $\mathfrak{m}$ is also finitely generated.
I am not sure if this is the kind of thing you are interested in, but let me at least state the easiest to prove criteria that I know for free-ness of $M$ in terms of vanishing of certain $\text{Ext}_R^i(\text{Tr}M, -)$ .
Proposition: If $M$ is a finitely generated module over a Noetherian local ring $R$ such that $M^*\ne 0$, and $\text{Ext}_R^{1,2}(\text{Tr} M, M^*)=0$, then $M$ is free. (See https://arxiv.org/abs/1805.04568 Lemma 4.12)
Proof: By definition, one have exact sequences $0\to M^*\to F\to K \to 0$ and $0\to K \to G\to \text{ Tr} M\to 0$ for some modules $K,F,G$, where $F$ and $G$ are free. Now $\text{Ext}_R^{1}(K, M^*)\cong \text{Ext}_R^{2}(\text{Tr} M, M^*)=0$, hence $0\to M^*\to F\to K \to 0$ splits, so $M^*$ is a free module. Since $M^*$ is non-zero, and now we know it is free, so $\text{Ext}_R^{1,2}(\text{Tr} M, M^*)=0$ yields $\text{Ext}_R^{1,2}(\text{Tr} M, R)=0$, hence $M$ is reflexive, so $M\cong M^{**}$ is free.
If you have other specific kind of situations in mind, I probably will be able to help you better ...
Best Answer
This is true for any $\mathrm{Ext}$-degree (and, in fact, without many hypotheses except that $\mathfrak m$ is finitely generated and $F$ is free).
Let $(x_1,\dots,x_n)$ be generators of the maximal ideal $\mathfrak m$. Then there is a surjection $G^n \to \mathfrak m G$ given by $$(g_1,\dots,g_n) \mapsto \sum x_i g_i.$$ Because $F$ is free, the map $F \to \mathfrak m G$ lifts to a map $F \to G^n$.
Now $\mathrm{Ext}(X,-)$ respects sums and takes the map multiplication-by-$x$ map $M \to M$ to the multiplication-by-$x$ map on $\mathrm{Ext}(X,M)$ (I'm assuming $R$ is commutative here, because otherwise $\mathrm{Ext}$ doesn't necessarily take values in $R$-modules). Therefore, the composite $$\mathrm{Ext}(X,G)^n \cong \mathrm{Ext}(X,G^n) \to \mathrm{Ext}(X,\mathfrak m G) \to \mathrm{Ext}(X,G)$$ is given by $(w_1,\dots,w_n) \mapsto \sum x_i w_i$, taking values in $\mathfrak m \mathrm{Ext}(X,G)$.
As a result, because the map $\mathrm{Ext}(X,F) \to \mathrm{Ext}(X,G)$ factors through this map $\mathrm{Ext}(X,G^n)$, it also factors through $\mathfrak m \mathrm{Ext}(X,G)$.