Let $p,q$ and $r$ be positive integers greater than $0$ with $q\neq r$. Suppose that $H$ is a finite connected graph without loops or multiedges on $p$ vertices with $q$ vertices of degree $r$, $r$ vertices of degree $q$ and $p =q+r$. I use the following notation: $V,E$ and $F$ are the number of points, edges and faces of $H$ respectively. By $deg(v)$ we mean the degree or valency of any vertex in $H$.
claim: $H$ has $qr$ edges
proof: If $V(r)$ and $V(q)$ are the set of vertices with degrees $r$ and $q$ respectively then $V(r) \cap V(q) = \emptyset$. Consequently $|V| = |V(r)\cup V(q)| = |V(r)| + |V(q)|$. Now, from Euler we know that $2|E| = \sum_{ v\in H} deg(v)$. In particular we can write
\begin{align} 2|E| = \sum_{v\in V(r)}deg(v)+\sum_{v\in V(q)}deg(v) =\underbrace{(r+\ldots+r)}_{q}+\underbrace{(q+\ldots+q)}_{r} = qr+rq =2rq. \end{align}
And that establishes the claim. $\blacksquare$
First I want to make sure that the above proof is correct. Note that $H$ is not necessarily bipartite! Then I have this question.
Question_1 For which pair $(q,r)$ can $H$ be planar. I suspect we have the following planar pairs: $(1,r)$, $(2,r)$, $(3,r)$,$(4,r)$ and $(5,r)$. In the case of $(5,r)$ I believe that $r=1,2,3$ or $4$.
There is no graph with $q=3$ and $r>8$.
If there was, then let $v_1,v_2,v_3$ be the three points of valence $r$ and $S$ be the $r$ points of valency $3$. Then how many points of $S$ are connected to only two of $v_1,v_2,v_3?$
The number of edges between $v_1,v_2,v_3$ and $S$ is at least $3(r-2)$. If $r>8$ then $3(r-2)>2r+2$. That means at least three nodes of $S$ must be connected to all three nodes of $v_1,v_2,v_3$. That means that we have $K_{3,3}$ as a subgraph.
If $(q,r)=(3,8)$, then $v_1,v_2,v_3$ have to all be joined, and there must be two $s_1,s_2\in S$ that must be joined to all $v_1,v_2,v_3$.
Every other node in $S$ must be joined with one other node in $S$ to get the valency of $3$.
Adding that to the fact that you've shown that $(q-3)(r-3)\leq 3$, and found cases where $(q,r)=(3,4),(4,5),(3,5),(4,6)$, you are left to search for $(q,r)=(3,6),(3,7)$ and $(3,8)$.
I think I've constructed a case where $(q,r)=(3,8)$. Make $v_1,v_2,v_3$ a triangle. Add $s_1$ to the center and connect to each of $v_1,v_2,v_3$. Add squares to the exterior of each triangle, drawing a complete $K_4$ for each square. Then add $s_2$ to the exterior and connect to each of the $v_1,v_2,v_3$.
I suspect this graph is unique.
You can get cases $(q,r)=(3,7)$ and $(3,6)$ by dropping one or both of $s_1,s_2,$ respectively.