Given a planar graph, embedded in the plane, you want to add a number of edges to get a maximal planar graph, and you want to be sure that resulting maximal planar graph won't have vertices with degree $=2$.
From this Wikipedia page:
A simple graph is called maximal planar if it is planar but adding any edge (on the given vertex set) would destroy that property...
The process of adding edges is straightforward - you need to find all faces, bounded by more than $3$ edges (including the external face), and triangulate these faces by adding new edges.
Let's consider the case, which looks problematic to you - a vertex $u$ with degree $2$, for which it's impossible to add a new edge, incident to the $u$. You have two vertices $v$ and $w$, adjacent to the vertex $u$, and two faces, partially bounded by edges $\{v,u\}$ and $\{u,w\}$. If a new edge, incident to the vertex $u$, can't be added, then both these faces should have an edge $\{v,w\}$, which prevents this addition. So, the graph contains two parallel edges $\{v,w\}$, which is not possible, because we consider simple graphs only.
Yes, they are (and well spotted).
Here's an old proof I have typed up. I think you can shorten it substantially however! To do so, I'd directly consider the walk bounding a face. It consists of alternating black / white vertices. Try argue using maximality that the walk is either a 4-cycle, or a star.
Lemma 1:
Every face of a 2-connected plane graph is bounded by a cycle.
This is well known (see, for example, Diestel)
Theorem 1:
Let $G$ be an mpb graph of minimum degree $\delta$ embedded in the plane. Then every face of $G$ is bounded by a 4-cycle if and only if $\delta \geq 2$.
Proof:
Let $G = (V,E)$ be an mpb graph and consider an embedding of $G$ in the plane. If every face of $G$ is bounded by a 4-cycle, then clearly $\delta \geq 2$. Conversely, assume $\delta \geq 2$. As $G$ is bipartite, $V$ can be partitioned into two sets, $B$ and $W$, such that every edge of $G$ is incident with one vertex of $B$ and one vertex of $W$.
First, we show that $G$ is 2-connected: Assume to the contrary, and without loss of generality, that $b_1\in B$ is a cut-vertex. Let $w_1, w_2$ be vertices of different components of $G-b_1$ such that $w_1, b_1, w_2$ are consecutive vertices of the walk bounding some face $f$ of $G$. As $w_1, w_2$ are adjacent to $b_1\in B$, it must be that $w_1, w_2 \in W$. Since the minimum degree of $G$ is at least 2, $w_2$ has a neighbour $b_2\in B$ such that $w_2, b_2$ are consecutive vertices of the walk bounding $f$. Since $b_2$ and $w_1$ are in different parts of the bipartition of $G$, and are both on the boundary of $f$, the maximality of $G$ implies that $w_1b_2 \in E$. Thus, $w_1, w_2$ lie in the same component of $G-b_1$, a contradiction. Hence $G$ is 2-connected.
We now show that every face of $G$ is a 4-cycle: Since $G$ is 2-connected and bipartite, by Lemma 1, every face is bounded by an even cycle. Thus it suffices to prove that every face of $G$ is bounded by a cycle of length less than 6. Assume, to the contrary, that some face $f$ of $G$ is bounded a cycle $C$ of length $k\geq 6$ and, without loss of generality, that $f$ is in the interior of $C$. Let $b_1, w_1, b_2, w_2, b_3, w_3$ be six consecutive vertices of $C$ with $b_1, b_2, b_3 \in B$ and $w_1, w_2, w_3 \in W$. As $b_1$ and $w_2$ lie on the same face and are in different parts of the bipartition of $G$, the maximality of $G$ implies that $b_1w_2 \in E$. Since the interior of $C$ is a face, $b_1w_2$ must lie in the exterior of $C$. Similarly to $b_1w_2$, the maximality of $G$ necessitates that $w_1b_3\in E$. If $w_1b_3$ lies inside $C$, it subdivides $f$, contradicting that $f$ is a face. If $w_1b_3$ lies outside $C$, it crosses $b_1w_2$, contradicting the planarity of $G$.
Theorem 2:
If $G$ is an mpb graph of minimum degree one, then $G$ is a star.
Proof:
Let $G = (V,E)$ be an mpb graph with partite sets $B$, $W$, and consider an embedding of $G$ in the plane. Without loss of generality, $v\in B$ is the neighbour of a vertex of degree 1. We claim that every neighbour of $v$ has degree 1. Assume to the contrary that $v$ has a neighbour of degree greater than 1. We can find neighbours $u,w$ of $v$ with $\text{deg}(u) = 1$ and $\text{deg}(w) \geq 2$ such that $u,v,w$ lie on the same face, $f$, of $G$. Let $b\in B$ be a neighbour of $w$, other than $v$, which lies on the boundary of $f$. Since $u$ and $b$ lie on the boundary of the same face, in different parts of the bipartition of $G$, they are adjacent by the maximality of $G$, contradicting the fact that the degree of $u$ is 1.
Best Answer
Taking the dual would give us a $4$-regular planar graph where we want as many of the faces as possible to be triangles. There's a family of $4$-regular polyhedra where almost all faces are triangles: the antiprisms. So the dual of an antiprism (apparently, this is called a trapezohedron) should be a good candidate for reaching the upper bound here.
For any even $n$, the dual of the antiprism consists of an $(n-2)$-cycle with two more vertices: one adjacent to every even vertex of the cycle, and one adjacent to every odd vertex. This has two vertices of degree $\frac n2-1$, and $n-2$ vertices of degree $3$. (When $n=8$, all $n$ vertices have degree $3$, and this graph is just the cube.)
Except when $n=8$, we cannot have $n$ vertices of degree $3$, but it's conceivable that we can have $n-1$ vertices of degree $3$ and one vertex of degree $n-5$. Let's analyze this case.
Let $v$ be the vertex of degree $n-5$. We know the bipartition, up to two cases: if we put $v$ on one side, and its $n-5$ neighbors on the other, then either the remaining $4$ vertices are all on the same side as $v$, or one of them is on the side with $v$'s neighbors and adjacent to the other $3$. In the first case, counting the edges from each side, we want $(n-5) + 4\cdot 3 = (n-5) \cdot 3$, so $n=11$. This is impossible by computer search, and there should be a combinatorial proof, but I haven't found a clean argument. In the second case, we want $(n-5) + 3 \cdot 3 = (n-4) \cdot 3$, so $n=8$, and that's the cube again.
One remaining question: can we have $n-2$ vertices of degree $3$ when $n$ is odd? Experimentally, no: here's a table of the maximum value of $n_3$ for small odd $n$.
\begin{array}{c|ccccccc} n & 11 & 13 & 15 & 17 & 19 & 21 \\ \hline n_3 & 8 & 10 & \color{red}{11} & 14 & 16 & \color{red}{17} \end{array}
For $n=15$ and $n=21$, we can't even get to $n-3$ vertices of degree $3$.
There's a reason to expect a repeating pattern mod $6$: if you can get a solution for $n$ where two vertices $v,w$ of degree greater than $3$ share a face but are not adjacent, you can get a solution for $n+6$ by adding a cube graph on $v$, $w$, and $6$ new vertices, and this can be iterated. It's hard to say what's up with the upper bounds, though.