The Euler identity for polyhedron:
can be transformed into this (see the details below):
and for graphs with all faces with degree 5 or 6, it becomes:
If you check the equilibrium of the identify, since $F_6$ face do not influence the equilibrium, a graph is forced to have 12 pentagonal faces and an indefinite number of $F_6$ faces.
As answered by Joseph Malkevitch, fullerenes have these characteristics.
$$F = F_2+F_3+F_4+\cdots$$
$$2E = 3V = 2F_2+3F_3+4F_4+\cdots$$
Since a region bounded by n edges has n vertices and each vertex belongs to three regions, by Euler's formula $V-E+F=2$ we have:
$$6V-6E+6F = 12$$
$$4E-6E+6F = 12$$
$$6F-2E = 12$$
$$6(F_2+F_3+F_4+\cdots)-(2F_2+3F_3+4F_4+\cdots) = 12$$
$$4F_2+3F_3+2F_4+F_5+0F_6-F_7-2F_8-3F_9-\cdots = 12$$