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.
Best Answer
Let $\{V_i\}$ be representatives from each of the isomorphism classes of simple left $S$-modules. For any finite length module ${}_S M$, let $\ell_S(M; V_i)$ denote the number of times that $V_i$ occurs in a composition series for $M$. Then the following formula holds (where almost all $\ell(M;V_i)$ are zero because $M$ has finite length, but any single $\ell_R(V_i)$ could be infinite, and $\infty \cdot 0 = 0$):$$\ell_R(M) = \sum \ell_R(V_i) \cdot \ell_S(M; V_i).$$
Thus, a finite length $S$-module $M$ has finite $R$-length if and only if every simple $S$-module that occurs in $M$ has finite $R$-length. This makes it clear that every finite length left $S$-module has finite $R$-length if and only if all simple left $S$-modules have finite $R$-length. The above discussion is true with no requirements on the ring $S$.
Now let's assume that $S$ is finitely generated as an $R$-module. (I'm not sure if this is what you meant by "finitely generated as an $R$-algebra," but it's probably true in the cases that you're studying if you're looking at maximal orders.) I'll prove that $S$ has finitely many simple modules, each of which has finite $R$-length. Let $J(A)$ denote the Jacobson radical of any ring $A$. Then $\mathfrak{m} = J(R) \subseteq J(S)$ because $S$ is a module-finite $R$-algebra; for a proof of this fact, see Lam's A First Course in Noncommutative Rings, Proposition 5.7. Now the simple $S$-modules are the same as the simple $S/J(S)$ modules (see Proposition 4.8 of the same text). But $S/\mathfrak{m} S \twoheadrightarrow S/J(S)$. Because $S$ is module-finite over $R$, $S/\mathfrak{m} S$ has finite $R$-length. Thus $S/J(S)$ has finite $R$-length. So $S/J(S)$ is artinian (the terminology here is that $S$ is a semilocal ring) and thus has finitely many simple modules, each of which has finite $R$-length. The same must be true for the simple $S$-modules
The formula above also verifies your formula in the case of a matrix ring. In case $S = \mathbb{M}_n(R)$, $S$ has a unique simple left module $V = (R/\mathfrak{m} \ \cdots \ R/\mathfrak{m})^T$, the set of all column vectors of length $n$ with entries in $R/\mathfrak{m}$. (At least one way to see that this is the only simple $S$-module is through Morita theory: the Morita equivalence between $R$-$\operatorname{Mod}$ and $S$-$\operatorname{Mod}$ sends the unique simple $R$-module $R/\mathfrak{m}$ to $V$, so that $V$ must be the unique simple $S$-module.) Since $\ell_R(V) = n$, the formula above reduces to $\ell_R(M) = n \cdot \ell_S(M)$.
(If something above doesn't make sense, you may want to review Jordan-Hoelder theory.)