This follows from the fact that for any prime $p$, and any integer $n\ge0$, we have $b_{n+p}\equiv b_{n+1} \pmod p$. This can be proved by a straightforward, though not very interesting, computation, using Fermat's theorem $2^p\equiv 2 \pmod p$ and the fact that $\binom {n+p}{i} \equiv \binom ni + \binom n{i-p} \pmod p$.

More interesting is a combinatorial proof. (For similar proofs and further references, see my paper Combinatorial proofs of congruences, in *Enumeration and Design*, ed. David M. Jackson and Scott A. Vanstone, Academic Press, Toronto, 1984, pp. 157–197, especially section 5.)

We consider the cyclic group $C_p$ acting naturally on $[p]=\{1,2,\dots, p\}$. This extends to an action on bicolored graphs with vertex set $[n+p]=\{1,2,\dots, n+p\}$; the group permutes vertices $1, 2, \dots, p$, and does not change the other vertices. Since every orbit has size 1 or $p$, $b_{n+p}$ is congruent modulo $p$ to the number of bicolored graphs on $[n+p]$ fixed by this action of $C_p$. It is not hard to see that these fixed bicolored graphs are those in which all vertices in $[p]$ have the same color, no edges join two vertices in $[p]$, and all vertices in $[p]$ have the same neighborhood. By "contracting" all the vertices in $[p]$ to a single vertex, whose neighborhood is the same as the neighborhood of any of the vertices in $[p]$, we get a bicolored graph on $n+1$ vertices. Thus the number of fixed bicolored graphs is $b_{n+1}$, so $b_{n+p}\equiv b_{n+1}\pmod p$.