What are the best known bounds on the number of non-isomorphic (unlabelled) planar graphs on $n$ vertices? Is there a simple proof that this number is at most exponential in $n$? 



Number of unlabelled planar graphs
Simple proof? Sort of. 


Let $t(n)$ be the number of plane triangulations with $n$ vertices. Tutte 
[in ''A Census of Planar Triangulations,'' Canad. J. Math. 14 (1962), 21-38] 
found an [explicit formula](https://www.cambridge.org/core/journals/canadian-journal-of-mathematics/article/census-of-planar-triangulations/0C07F660D1000D2AF1DA9065E84FA3E4) for $t(n)$, implying that $t(n)\sim c n^\alpha \gamma^n$, where $\gamma=256/27$. From here, the number $a(n)$ of non-isomorphic planar graphs is at most $2^{3n-6}t(n)$, also exponential. There is an elegant [bijective proof](http://www.lix.polytechnique.fr/~poulalho/Articles/PoSc_Algorithmica06.pdf) of Tutte's formula by Poulalhon and Schaeﬀer, but I wouldn't call it "simple". 


For an exponential upper bound, there is a way to simplify the Poulalhon-Schaeﬀer proof. Take a rooted plane triangulation $G$ (rooted means a vertex and an adjacent edge are labeled). Denote by $T$ the depth-first search tree which is a planar tree with $n$ vertices and the same root. The number of such trees is the Catalan number $C\_{n-1}$. Now we need to triangulate the complement of $T$ in the plane, which is bounded by the number of triangulations of $2(n-1)$-gon (twice the number of edges of $T$). The latter is equal to $C\_{2n-4}$. Since $C\_k<4^k$, putting these together gives an upper bound
$$a(n) < 2^{3n} \cdot 4^n \cdot 4^{2n} = 2^{9n}.$$
One can easily get rid of the $2^{3n}$ term by realizing that you are overcounting, but that requires a bit of an explanation and a tiny bit more background. 


