I've looked a "local optima", i.e., sets of squares whose number of tilings is larger than any set of squares obtained by moving one or two squares. For small $N$ only the global optimum is a local optimum, but for 6 dominoes, the (two possible) configuration(s) made by gluing a 2×2 square to the long side of a 2×4 rectangle are local optima with 12 tilings (compared to the 13 tilings of 3×4). This suggests that a local approach would not work so well.

For small values of $N$, shapes with $2N$ squares which maximize the number of tilings by $N$ dominoes seem to be the following rectangles: 2×1, 2×2, 2×3, 2×4, 2×5, 2×6 (not the squarer 3×4). But 4×4 (36 tilings) is better than 2×8 (34 tilings), consistent with your conjecture.

I've now read Fiore and Leinster. For a given relation $T=p(T)$, they define an order $q_1\leq q_2$ on polynomials. If $p$ is nice enough (in particular, of degree $\geq 2$ and with non-zero constant coefficient), then $q_1\leq q_2$ (and $q_2\leq q_1$) for any non-constant $q_i$. I just want to point out that their order is the same as what I hint to at the end of my answer: $q_1\leq q_2$ if $\exists C,\forall n$ the $n$-th coefficient of $q_1(T(z))$ is smaller than the $(n+C)$-th coefficient of $q_2(T(z))$. For the Fibonacci/geometric cases, $p$ is linear and their theorem does not apply.

In fact, the coincidence that $1+2+4+8+\dots=-1=1+1+2+3+5+8+13+\dots$ comes from a bijection between two descriptions of the set of finite sequences of 1 and 2. There are $2^n$ sequences of $n$ digits, so $1+2+4+\dots$ sequences. There are $f_n$ sequences which sum to $n$, so $1+1+2+3+5+\dots$ sequences.

A (counter-)example with $S^2$ topology. Let OABCD be a square pyramid and let A' and C' be the mid-points of OA and OC. The result is a (degenerate) polyhedron with five quadrilateral faces, and the points OABCD have the required property. If you don't allow collinear vertices then shift A' and C' a bit, keeping OA'AD and OC'CB flat, then replace the no-longer-flat faces OA'AB and OC'CD by five quadrilaterals each (glue a "cube" onto each face).

The conjecture is too general (if I understand it correctly). Take a 3×3 grid of squares and identify opposite edges to make a polyhedron with square faces and the topology of the torus. Any two points belong to a common facet, so the maximum number of vertices such that [...] is 9. The same happens for a periodic grid of 3×…×3 hypercubes.

Isn't it enough to add $0$ to the set to get a $B_3[1]$ set? (Bearing in mind I have not looked up what $B_3[1]$ means...)