Do there exist "genuinely maximal" theories $\text{Th}(\mathcal M)$? Given a language $\mathcal L$ and an $\mathcal L$-structure $\mathcal M$, the theory of $\text{Th}(\mathcal M)$ is defined to be the set of $\mathcal L$-sentences $\phi$ for which $\mathcal M \vDash \phi$. Such theories are maximal, meaning that for each $\mathcal L$-sentence $\phi$ either $\phi \in \text{Th}(\mathcal M)$ or $\neg \phi \in \text{Th}(\mathcal M)$. This is because either $\mathcal M \vDash \phi$ or $\mathcal M \nvDash \phi$, simply by merit of the interpretation function $\cdot^\mathcal M : \text{Formula}(\mathcal L) \to \{T,F\}$, uh, existing.
Although theories of $\mathcal L$-structures "are maximal", this fact may not be realizable. By this I mean the following: there may exist an $\mathcal L$-sentence $\phi$ for which, despite that $\phi \in \text{Th}(\mathcal M) \lor \neg \phi \in \text{Th}(\mathcal M)$, it is the case that $\phi \in \text{Th}(\mathcal M)$ is not provable and $\neg \phi \in \text{Th}(\mathcal M)$ also is not provable.
Let me give a specific example. For this example I assume that $\Phi$ is some undecidable sentence within which the system we are currently working; for instance, assuming we are working in ZFC (within which we have encoded model theory), we could have $\Phi$ be $2^{\aleph_0} = \aleph_1$.
Let $\mathcal L = \{S, z\}$ be the language of the natural numbers, and $\Gamma$ the axioms of Robinson arithmetic. Define a model $\mathcal M$ for $\Gamma$ as follows: if $\Phi$ is false, then let $\mathcal M = (\mathbb N;\ 0, +)$ be the standard model of the naturals; if $\Phi$ is true, then let $\mathcal M = (\mathbb N \cup \{\infty\};\ 0, +)$ be the standard model plus an additional element $\infty$ for which $S(\infty) = \infty$.
Note that in this example we still have that $\text{Th}(\mathcal M)$ "is complete", but not in any real way. Letting $\phi$ be the sentence $\exists i : S(i) = i$, we have since $\mathcal M$ is well-defined that either $\mathcal M \vDash \phi$ or $\mathcal M \vDash \neg \phi$, but in fact it is impossible to prove either way due to undecidability of $\Phi$.
My questions are these:

*

*Are there any examples of this effect that are less "gimmicky"? In my example I basically shoehorn in a poorly-behaved $\phi$ by assuming existence of an undecidable $\Phi$. Is there an example where the $\phi$ arises "naturally" from the model, not requiring existence of an external undecidable $\Phi$?


*Are there any ((meta-)meta-)theorems along these lines? For instance, here's a conjecture: if $\Gamma$ is strong enough for Goedel's incompleteness theorem to apply, and if $\mathcal M \vDash \Gamma$, then there exists a sentence $\phi$ for which neither $\mathcal M \vDash \phi$ nor $\mathcal M \nvDash \phi$ are provable, despite their disjunction being true.
(I have a small handful of other examples of this effect, but they are similarly gimmicky. I also tried my hand at proving this conjecture, but eventually concluded that my method is likely flawed.)
 A: As often happens with logic, there are some subtleties around precisely posing the question you intend to ask. Let me rephrase it as follows:
Fix a "background theory" to do model theory in - say, $\mathsf{ZFC}$. A definition of a structure will be a formula $\varphi$ in the language of set theory such that $\mathsf{ZFC}\vdash$ "$\varphi$ defines a structure" (note that part of this is that $\mathsf{ZFC}\vdash\exists! x\varphi(x)$). When $\varphi$ is such a formula, I'll write "$\mathcal{M}_\varphi$" for the structure defined by $\varphi$. Note that different background models of $\mathsf{ZFC}$ may compute $\mathcal{M}_\varphi$ differently, but for any fixed $W\models\mathsf{ZFC}$ there will be a unique-and-well-defined value of $\mathcal{M}_\varphi^W$ ("what $W$ thinks $\varphi$ defines"). Similarly, let $\Sigma_\varphi$ be the language of $\mathcal{M}_\varphi$; for simplicity, let's only look at situations where $\Sigma_\varphi$ is finite and "fully pinned down" by the $\mathsf{ZFC}$ axioms (this just makes things somewhat easier to think about).
Now say that a definition of a structure $\varphi$ is bivalent iff for every $\Sigma_\varphi$-sentence $\theta$ we have $$\mathsf{ZFC}\vdash\theta^{\mathcal{M}_\varphi}\quad\mbox{or}\quad\mathsf{ZFC}\vdash\neg\theta^{\mathcal{M}_\varphi}.$$ You're essentially asking about non-bivalent definitions of structures.
One quick observation is that, as you suspect, any $\varphi$ which $\mathsf{ZFC}$-consistently defines a structure rich enough to implement arithmetic must be non-bivalent. This is because we can encode the Godel-Rosser sentence for $\mathsf{ZFC}$, or indeed any c.e. extension of $\mathsf{ZFC}$, as a $\Sigma_\varphi$-sentence. In particular, if we let $\varphi$ be the usual definition of the natural numbers with addition and multiplication inside set theory, then $\varphi$ is non-bivalent. Indeed, Godel's incompleteness theorem shows that this remains the case no matter how we strengthen $\mathsf{ZFC}$, as long as we remain consistent and c.e.
On the other hand, bivalent definitions of structures also exist! The usual definition of the field of real numbers is an example; see this old answer of mine.
Now on to the topic of gimmicks. Informally, you've shown that every structure which has a bivalent definition also has a non-bivalent definition; a bit more precisely, if $\varphi$ is a bivalent definition of a structure, then letting $\theta$ be a sentence independent of $\mathsf{ZFC}$ there are non-bivalent definitions of structures $\varphi_1,\varphi_2$ such that $\mathsf{ZFC}$ proves that $\varphi_1$ coincides with $\varphi$ iff $\theta$ and $\varphi_2$ coincides with $\varphi$ iff $\neg\theta$. This raises the natural question of whether there is a structure which has "copies" definable in every model $\mathsf{ZFC}$ but has no bivalent definition:

Is there a structure $\mathcal{A}$ such that

*

*for every $W\models\mathsf{ZFC}$, there is some definition of a structure $\varphi$ with $\varphi^W\cong \mathcal{A}$, but


*for every bivalent definition of a structure $\psi$, there is some $W\models\mathsf{ZFC}$ with $\psi^W\not\cong\mathcal{A}$?

I believe the answer to this is no, and I have a sketch of an argument, but I don't have time to flesh it out right now; I'll add it later today if it holds up, and if it doesn't hold up I'll explain why it breaks.
