The friendship theorem states that if every pair of people has exactly one common friend, then one person is friends with everybody. Which is unfortunately not easy to actually proof. I came up with the following variation on this theorem regarding bipartite graphs:
if every pair of females has exactly one mutual male friend, and every pair of males has exactly one mutual female friend (the bipartite friendship condition), but the amount of females and males are not equal, then one male is friends with every girl and one girl is friends with every boy.
If the amount of females and males are equal and more than 3, then there is another possible configuration: there is a popular male and popular female, who are friends with everyone from the other gender , except with each other. Everyone else has exactly 2 friends of opposite gender. 
Can we prove that these are the only bipartite graphs satisfying the bipartite friendship condition?
Think of the two sides of the graph instead as being "points" and "lines", with edges of the graph representing the incidence relation (lines passing through points). Then the friendship condition becomes:
Of course, these are not necessarily geometric lines, but just a metaphor. (Incidence geometry studies the abstract objects represented by points and lines without a geometric meaning.)
The De Bruijn–Erdős theorem narrows things down. It tells us three things:
We want both sets of assumptions, so we want both conclusions, and so there must be an equal number of points and lines. De Bruijn and Erdős described the equality case. There are two possibilities: first, we can see a "near pencil", which is a line with $n-1$ out of $n$ points on it, and an $n^{\text{th}}$ point with $n-1$ lines joining it to the other $n-1$ points. This is a geometric description of the second construction in the question.
The other equality case comes from finite projective planes. Here, there must be $n = q^2+q+1$ points and $n = q^2+q+1$ lines, for some integer $q$ called the "order" of the projective plane. Each point lines on $q+1$ lines, each line passes through $q+1$ points, and (as desired) any two points lie on a unique line, and any two lines have a unique intersection.
Projective planes are known to exist when $q$ is a prime power, though we haven't ruled out the possibility that they exist in other cases. The simplest new construction this gives us is the Fano plane for $q=2$:
This can be turned into a bipartite graph with $7$ vertices on each side, where each vertex has $3$ neighbors on the other side, satisfying the condition in the problem.
Though there are infinitely many projective planes that answer the question, I must also point out that they are all symmetric: the number of points equals the number of lines. This gives a balanced bipartite graph where both sides have an equal number of vertices.
By the converse of our application of De Bruijn–Erdős, if we have an unbalanced bipartite graph, the only possible construction is the "boring" one: one vertex from each side is adjacent to every vertex on the other side, and all other vertices have degree $1$.