The analogy doesn't quite give a number theoretic version of the Poincare conjecture. See Sikora, "Analogies between group actions on 3-manifolds and number fields"
(arXiv:0107210): the author states the Poincare conjecture as "S3 is the only closed 3-manifold with no unbranched covers." The analogous statement in number theory is that Q is the only number field with no unramified extensions, and indeed he points out that there are a few known counterexamples, such as the imaginary quadratic fields with class number 1.
The paper also has a nice but short summary of the so-called "MKR dictionary" relating 3-manifolds to number fields in section 2. Morishita's expository article on the subject, arXiv:0904.3399, has more to say about what knot complements, meridians and longitudes, knot groups, etc. are, but I don't think there's an explanation of what knot surgery would be and so I'm not sure how Kirby calculus fits into the picture.
Edit: An article by B. Morin on Sikora's dictionary (and how it relates to Lichtenbaum's cohomology, p. 28): "he has given proofs of his results which are very different in the arithmetic and in the topological case. In this paper, we show how to provide a unified approach to the results in the two cases. For this we introduce an equivariant cohomology which satisfies a localization theorem. In particular, we obtain a satisfactory explanation for the coincidences between Sikora's formulas which leads us to clarify and to extend the dictionary of arithmetic topology."
This is a great question. Someone will come along with a better answer I'm sure, but here's a bit off the top of my head:
1) The Hilbert class field of a number field $K$ is the maximal everywhere unramified abelian extension of $K$. (Here when we say "$K$" we really mean "$\mathbb{Z}_K$", the ring of integers. That's important in the language of etale maps, because any finite separable field extension is etale.)
In the case of a curve over $\mathbb{C}$, the "problem" is that there are infinitely many unramified abelian extensions. Indeed, Galois group of such is the abelianization of the fundamental group, which is free abelian of rank $2g$ ($g$ = genus of the curve). Let me call this group G.
This implies that the covering space of C corresponding to G has infinite degree, so is a non-algebraic Riemann surface. In fact, I have never really thought about what it looks like. It's fundamental group is the commutator subgroup of the fundamental group of C, which I believe is a free group of infinite rank. I don't think the field of meromorphic functions on this guy is what you want.
2) On the other hand, the Hilbert class group $G$ of $K$ can be viewed as the Picard group of $\mathbb{Z}_K$, which classifies line bundles on $\mathbb{Z}_K$. This generalizes nicely: the Picard group of $C$ is an exension of $\mathbb{Z}$ by a $g$-dimensional complex torus $J(C)$, which has exactly the same abelian fundamental group as $C$ does: indeed their first homology groups are canonically isomorphic. $J(C)$ is called the Jacobian of $C$.
3) It is known that every finite unramified abelian covering of $C$ arises by pulling back an isogeny from $J(C)$.
So there are reasonable claims for calling either $G \cong \mathbb{Z}^{2g}$ and $J(C)$ the Hilbert class group of $C$. These two groups are -- canonically, though I didn't explain why -- Pontrjagin dual to each other, whereas a finite abelian group is (non-canonically) self-Pontrjagin dual. [This suggests I may have done something slightly wrong above.]
As to what the Hilbert class field should be, the analogy doesn't seem so precise. Proceeding most literally you might take the direct limit of the function fields of all of the unramified abelian extensions of $C$, but that doesn't look like such a nice field.
Finally, let me note that things work out much more closely if you replace $\mathbb{C}$ with a finite field $\mathbb{F}_q$. Then the Hilbert class field of the function field of that curve is a finite abelian extension field whose Galois group is isomorphic to $J(C)(\mathbb{F}_q)$, the (finite!) group of $\mathbb{F}_q$-rational points on the Jacobian.
Best Answer
For this analogy, like most analogies in mathematics, and indeed like most philosophical principles in mathematics, one can certainly make a part of it formal and rigorous, but I don't think any true formal statement could ever capture all of what we mean by the analogy.
In particular, by well-chosen definitions, one can write down statements of the form "If X is either a 3-manifold or the ring of integers of a number field, then something is true about X", where "something" is expressed the same way in each case. But there's no reason to expect that there is a single statement that implies all true such statements.
In particular, one can certainly not get an equivalence of categories between some category of 3-manifolds and some category of number fields (as Wojowu suggests in the comments), or any kind of correspondence between one 3-manifold and one number field, that respects the interesting structure like Artin-Verdier duality. (Thus I think the equivalence of fundamental groups between $S^3$ and $\mathbb Z$ is a red herring.)
Note that in Verdier duality, a pretty fundamental concept is an orientation. Any 3-manifold is either oriented or has a double cover to be oriented. But from the form of Artin-Verdier duality, for a number field to be oriented, it would have to contain the $n$th roots of unity for all $n$, which is impossible. So the "oriented double cover" in this setting is actually a cover of infinite degree! Covering spaces and dualizing sheaves are some of the concepts we absolutely do want to match up, so I don't think there's any way to wriggle out of this.