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 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 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.

There is a short proof of a stronger result by Norine, Seymour, Thomas, and Wollan that there is an exponential upper bound for every proper minor-closed class.

For every proper minor-closed class of graphs $\mathcal G$, there exists a constant $c$, such that for every integer $n$, there are at most $n! c^n$ graphs in $\mathcal G$ with vertex set $\{1, \dots, n\}$.

This is the labelled version of your question, but as Igor mentions, the typical graph has trivial automorphism group, so we can divide by $n!$.

The paper is entitled Proper minor-closed families are small and can be downloaded here or from Robin Thomas's webpage.