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](https://doi.org/10.1016/B978-0-7204-2233-7.50025-3)_ 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](https://mathoverflow.net/a/350378/8133).

*Below, “definable” means “definable without parameters.” For more pleasant language I'll call this argument the "Tarskian argument."*

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## 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.

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## 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 the article _[Abstract Computability and Invariant Definability](https://doi.org/10.2307/2270854)_ 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.