A family $F$ of graphs on a common vertex set is triangle-intersecting if every pair of graphs in $F$ shares at least one triangle, i.e. $G_1 \cap G_2$ contains a 3-clique.
Instances For
Any graph on at most 2 vertices is triangle-free (3-clique-free).
Base case of the triangle-intersecting bound for $n \le 2$: such a family must be empty.
A graph on 3 vertices that contains a 3-clique must be the complete graph $K_3$.
The triangle-intersecting bound for $n = 3$: the family contains only $K_3$, so its cardinality is at most $1 < 2$.
From the Shearer-style power inequality $|F|^k \le (2^{r-1})^s$ combined with the arithmetic gap $(r-1) s < (m-2) k$, deduce the desired bound $|F| < 2^{m-2}$.
Existence of a Shearer-style bipartition covering for any non-empty triangle-intersecting family on $n \ge 4$ vertices, providing parameters $r,s,k$ along with the projection bound $|F|^k \le (2^{r-1})^s$ used to derive the main theorem.
Chung-Graham-Frankl-Shearer (Theorem 10.4.9): every triangle-intersecting family of graphs on $n$ labeled vertices has size strictly less than $2^{\binom{n}{2} - 2}$.