There’s also an “invariant definability” argument. I’ll sketch it quickly below, and then give an analysis to explain why I think it’s meaningfully different. Embarrassingly I can't find a source for it at the moment; I recall seeing it as a footnote in Kreisel's model-theoretic invariants paper, but it doesn't seem to be there. Multiple authors have written on invariant definability (which this answer is not-so-secretly an advertisement of) so I haven't yet been able to conduct an exhaustive search for the reference, but when I find it I'll update this. Incidentally, this argument was referred to at the beginning of another answer of mine.
Below, “definable” means “definable without parameters.” For more pleasant language I'll call this argument the "Tarskian argument."
##Argument
Let $T$ be an “appropriate” theory of arithmetic (say, $T\supseteq R$). We tweak Tarski’s undefinability theorem very slightly as follows. For $X\subseteq\mathbb{N}$ and $M\models T$, say that $X$ is pseudo-definable in $M$ if $X=D\cap \mathbb{N}$ for some definable $D\subseteq M$. We then have:
$(*)\quad$ Suppose $M\models T$. Then $Th(M)$ is not pseudo-definable in $M$.
The standard proof still works: supposing to the contrary that $\theta$ pseudo-defined $Th(M)$ in $M$, let $m$ be the Godel number of the formula $\eta(x)\equiv$ "$\theta$ fails on the number of the sentence gotten by plugging $x$ into the formula with number $x$" - appropriately formalized - and consider the sentence $\eta(\underline{m})$ (applying representability appropriately).
With this in hand we argue as follows. Suppose $S\supseteq T$ is computable and satisfiable. Then by representability we have that $S$ is pseudo-definable in $M$ for every $M\models T$. Taking $M\models S$, we have by $(*)$ that $S\not=Th(M)$, so $S$ is not complete.
##Analysis
Now let me argue in favor of the Tarskian argument being a genuine variation.
First, there’s an easy negative observation: it applies to Willard’s self-verifying theories and so cannot yield the second incompleteness theorem as a direct corollary. More subjectively, the argument is fairly non-constructive, and doesn’t (as far as I can see) quickly yield a specific undecidable sentence.
Of course, this is also a feature of the many standard computability-theoretic arguments. I think there’s still a difference here - this time a positive one - due to the way the Tarskian argument interacts with the notion of invariant definability. There are a couple ways to frame this - see e.g. the beginning of this article by Moschovakis for some discussion - and I'll use the following:
Definition:
- An arithmetic context is a set $\mathfrak{C}$ of models of Robinson's arithmetic $R$.
- For an arithmetic context $\mathfrak{C}$, a set $A\subseteq \mathbb{N}$ is $\mathfrak{C}$-invariantly definable if there is some formula $\varphi$ with $\varphi^M\cap\mathbb{N}=A$ for all $M\in\mathfrak{C}$.
(Here I’m indulging in the usual abusive conflation of $\underline{k}^M$ and $k$.)
- For a theory $E$ and an arithmetic context $\mathfrak{C}$, say that $E$ is $\mathfrak{C}$-satisfiable if some member of $\mathfrak{C}$ satisfies $E$.
Then the Tarskian argument in fact gives:
Proposition: Suppose $\mathfrak{C}$ is an arithmetic context. Then no $\mathfrak{C}$-satisfiable theory is $\mathfrak{C}$-invariantly definable.
(Since the computable sets are invariantly definable in every arithmetic context and every theory extending $R$ yields an arithmetic context, this is a generalization of essential undecidability.)
The point is that in general the $\mathfrak{C}$-invariantly definable sets need not support a good computability theory in any sense: by judiciously terrible choice of $\mathfrak{C}$, we can make the set of $\mathfrak{C}$-invariantly definable sets have basically no structural properties besides forming a Boolean algebra. So, for example, I don’t see how to whip up an analogue of the “inseparable c.e. sets” argument for arbitrary arithmetic contexts.
Of course, pathological arithmetic contexts are uninteresting, so it’s hard to argue that this aspect is actually valuable in any way. But it is - as far as I can tell - a nontrivial feature.