The same way we can take n to be countably inf and can prove via diagnolization that 2^n must be uncountable, is there the same style of argument for constructing a countably infinite set? An argument that goes 2^(something) is considered countably inf.
My assumption is the "something" can't be finite--2^(any finite number) is going to be finite--this is why the title of the question.
No: there is no cardinal $\kappa$ such that $2^\kappa=\omega$. $2^n$ is finite for all finite $n$, and $2^\kappa$ is uncountable for all infinite $\kappa$. This really isn’t very strange: even in the non-negative integers the powers of $2$ have gaps, and only some integers are powers of $2$.
An interesting question is what gaps there are amongst the infinite cardinalities. For instance, is $2^\omega$ equal to $\omega_1$, the first uncountable cardinal, or is it bigger, so that $\omega_1$ is not a power of $2$ at all? To put it a bit differently, is there an infinite set that is uncountable but strictly smaller in cardinality than $\Bbb R$? It turns out that the usual axioms of set theory aren’t strong enough to answer this question: it is consistent with them that $2^\omega=\omega_1$, so that there is no such set, and it is also consistent with them that $\omega_1<2^\omega$, so that there is such a set. The assertion that there is no such set is known as the continuum hypothesis (CH) and has a long history. The generalized continuum hypothesis (GCH), that if $\kappa$ is any infinite cardinal there is no cardinal strictly between $\kappa$ and $2^\kappa$, has also been around for over a century and is also consistent with but independent of the usual axioms for set theory. But even if we assume the GCH, there are still uncountable cardinals that are not powers of $2$, starting with $\omega_\omega$, the $\omega$-th infinite cardinal.