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
Yes.
Start with a maximal planar graph that already has this property (for example, $K_4$ trivially has this property).
Now you subdivide each edge $uv$ twice, adding degree 2 vertices $x$ and $y$. Further, add a vertex $w$ to every face of the planar graph. You will now add every edge of the form $wx$ and $wy$ (leaving $w$ with degree 6, and $x$, $y$ with degree 5). You will also add edges between the
$x$' vertices on the same face, and the$y$' vertices on the same face to get these up to degree 6.This should work for any starting graph to add only vertices of degree 6.
See the diagram below for how to do this step by step on $K_4$:
The blue vertices are $w$ vertices, red vertices are $x$ and $y$ vertices. There are 6 blue edges for each $w$ vertex, and 2 original black edges, 2 blue edges, 2 red edges for each edge among $x$ and $y$ type vertices.
EDIT: Of course, to get the entire infinite class, you repeat this construction again and again on the previously obtained graph.