Actually the requirement that your graph is vertex transitive is *very* strong. Any such tiling is "close" to the Cayley graph of a hyperbolic surface group, i.e. the fundamental group of a closed surface, which is not a sphere, torus, projective plane, or Klein bottle. Unfortunately the best description I can give you is algorithmic.

First of all let's assume that the tiling doesn't have any ideal polygons. I am presuming that you are interested in Cayley graphs of finitely generated groups, so the graph should be locally finite. Also every vertex should be adjacent to an even number of edges.

Let $\Gamma$ be the automorphism group of some tiling $\mathcal T \subset \mathbb{H}^2$ without ideal polygons. Any $g \in \Gamma$ that fixes a point must be of finite order, therefore $\Gamma$ acts properly discontinuously and cocompactly by isometries on $\mathbb{H}^2$, and $\Gamma$ is the fundamental group of a hyperbolic orbifold $\mathcal{O}$ without boundary.

If there is a group $H\leq \Gamma$ that acts *freely* **and** *vertex transitively* of $\mathcal T$, then $\mathcal T$ is a Cayley graph of $H$, which is the fundamental group of a closed surface. This $H$ would be the subgroup corresponding to some "good" (technical term signifying a manifold) cover of $\mathcal{O}$ of minimal degree. In particular $[\Gamma: H]<\infty$. Although any automorphism group $\Gamma$ has "good" subgroups $H$ that act freely on the tiling, it may be that none of these subgroups are vertex transitive.

It follows that if a tiling $\mathcal T$ that is the Cayley graph of a group, then there must exist some closed hyperbolic surface $\Sigma$ that can be decomposed as a union of polygons with a single vertex (i.e. all the corners of the polygons are at the same point.) Call this tiling of $\Sigma$ $\mathcal P$. The tiling $\mathcal T$ is obtained as the universal cover of $\mathcal P$.

Since the number of edges that meet at a vertex of $\mathcal T$ give an upper bound for the rank of $\pi_1(\Sigma)$ this bounds the number of surfaces to check, so your property can be recognized algorithmically.

To answer your questions about triangles, an Euler characteristic of surfaces calculation gives us $n = 3F$ where $F\geq 4$ is any even number.