Let $\mathfrak G$ be a class of graphs with quadratically many maximal cliques in the size of the vertex set. In other words, if $G=(V,E)\in \mathfrak G$ with $|V|=n$, the graph $G$ contains $O(n^2)$ maximal cliques.
Can $G$ be realized as the $1$-skeleton of a Vietoris-Rips complex $Rips(X)_r$ for some finite $X\subset \mathbb R^2$ (with the $L_2$ metric) at some scale parameter $r \geq 0$?
Recall that the Vietoris Rips complex on a finite set $X$ in a metric space at a scale parameter $r$ is the maximal simplicial complex defined by saying that a subset $S\subset X$ of $k$ points in $X$ is a $(k+1)$-simplex whenever $\text{diam} (S)\leq r$.
The Rips complex is completely determined by its $1$-skeleton, making it a clique complex.
I know that any planar or $1$-planar graph has only $O(n)$ maximal cliques, so therefore $\mathfrak G$ must not be contained in these classes, though it could potentially be $k$-planar for $k\geq 2$.
I've calculated numerous simple examples in $\mathbb R^2$ and I can't get quadratically many maximal cliques.
It seems that the geometry of $\mathbb R^2$ restricts this, but I'm not completely sure how to prove it rigorously.
Remark (edit): In full generality this question could be stated for classes of graphs having $O(n^\ell)$ maximal cliques for $\ell\geq 2$, but it seemed like asking when $\ell=2$ would be the simplest case to show.