An addition to Douglas S. Stones' post, who mentioned $K_{3,3}$. This graph is bipartite and the shortest length of cycle in this graph is $4$.
For $K_5$ and $K_{3,3}$ the following criteria, which are mentioned in the Wikipedia article on planar graphs, can be used. (Those are precisely the results which you refer to as Theorem 1 and Theorem 2 in your post.)
For a planar graph having $v$ vertices and $e$ edges, the following holds:
- If $v \ge 3$ then $e \le 3v - 6$;
- If $v \ge 3$ and there are no cycles of length $3$, then $e \le 2v - 4$.
(The first one can be used to show that $K_5$ is not planar, the second one can be used for $K_{3,3}$.)
The idea of both proof is similar. And it is based on Euler's formula $v-e+f=2,$ which is equivalent to $f=2-v+e$.
Boundary of each face consists of at least $3$ edges. Since every edge belongs to boundaries of two faces, if you add together the lengths of boundaries of all faces, you get precisely $2e$. So we see that
$2e\ge 3f$
$2e\ge 6-3v+3e$
$3v-6\ge e$.
If we, moreover, know that the shortest cycle as at least of length $4$, we get that each boundary must have at least $4$ edges. Hence
$2e\ge 4f$
$2e\ge 8-4v+4e$
$e\ge 4-2v+2e$
$2v-4\ge e$.