I know almost nothing about noncommutative rings, but I have thought a bit about what the general concept of spectra might or should be, so I'll venture an answer.
One other property you might ask for is that it has a good categorical description. I'll explain what I mean.
The spectrum of a commutative ring can be described as follows. (I'll just describe its underlying set, not its topology or structure sheaf.) We have the category CRing of commutative rings, and the full subcategory Field of fields. Given a commutative ring $A$, we get a new category $A/$Field: an object is a field $k$ together with a homomorphism $A \to k$, and a morphism is a commutative triangle. The set of connected-components of this category $A/$Field is $\mathrm{Spec} A$.
There's a conceptual story here. Suppose we think instead about algebraic topology. Topologists (except "general" or "point-set" topologists) are keen on looking at spaces from the point of view of Euclidean space. For example, a basic thought of homotopy theory is that you probe a space by looking at the paths in it, i.e. the maps from $[0, 1]$ to it. We have the category Top of all topological spaces, and the subcategory Δ consisting of the standard topological simplices $\Delta^n$ and the various face and degeneracy maps between them. For each topological space $A$ we get a new category Δ$/A$, in which an object is a simplex in $A$ (that is, an object $\Delta^n$ of Δ together with a map $\Delta^n \to A$) and a morphism is a commutative triangle. This new category is basically the singular simplicial set of $A$, lightly disguised.
There are some differences between the two situations: the directions have been reversed (for the usual algebra/geometry duality reasons), and in the topological case, taking the set of connected-components of the category wouldn't be a vastly interesting thing to do. But the point is this: in the topological case, the category Δ$/A$ encapsulates
how $A$ looks from the point of view of simplices.
In the algebraic case, the category $A/$Field encapsulates
how $A$ looks from the point of view of fields.
$\mathrm{Spec} A$ is the set of connected-components of this category, and so gives partial information about how $A$ looks from the point of view of fields.
Can you show, from whatever you have available to you in the course at this point, that maximal ideals have inverses (as $A$-modules)? It would then follow that if a max. ideal $\mathfrak m$ contains an ideal $\mathfrak a$ then $\mathfrak m$ is a factor of $\mathfrak a$. Then maybe by a Noetherian inductive process show any primary ideal is a prime power by starting with a primary ideal, picking a maximal ideal containing it (we know secretly there's only 1 choice), write your primary ideal as a product of that maximal ideal and another ideal, show that other ideal is primary, and repeat. At the end you could read off that all the maximal ideal factors must be the same. I'm not saying I have worked out the details on that, but it's still not completely clear to me what is known and not known to the students, so this is just a suggestion.
I looked at my own lecture notes to see how I showed a maximal ideal in a Dedekind domain has an inverse, and I used a variant on the determinant trick together with the ring being integrally closed. I vote for using the determinant trick. Then you'd use it now once and you have already told us that you will use it again later. After you use it twice, it becomes a method and not a trick. :) Let's think about it this way: you are willing to use the raw definition of being integrally closed, and you certainly need to use integral closedness somewhere, but if you want to produce a monic polynomial at some point so you can use the fact that your ring $A$ is integrally closed, where in the world are you going to get such polynomials from? The determinant trick is one way. What other way is there in this proof?
I am not sure that any solution you eventually find will be satisfying to the students, to whom this is being seen for the first time. The next time you teach the course, consider covering the material in a different order so you don't get caught in the same way.
Best Answer
You could try having a look at Yekutieli and Zhang's paper Homological Transcendence Degree (http://arxiv.org/abs/math/04120130). They call a $k$-algebra $A$ "doubly Noetherian" if $A \otimes_k A^{op}$ is Noetherian, and "rationally Noetherian" if $A \otimes_k U$ is Noetherian for every division ring $U$.
There's a lot of other stuff in the paper too, but the first couple of sections look at several conditions for when rings (usually simple Artinian or division) are doubly or rationally noetherian, and there's a nice example of a field in section 7 which is not doubly noetherian.
It's not quite what you asked (it certainly won't help with the $q$-plane, but maybe the $q$-torus or the $q$-division ring?) but it might help with intuition and provide some examples.