The following is a less sophisticated version of the method of Thomas Andrews. We will count the complement, the strings that have no pair of consecutive $0$s. Then we can subtract this number from $256$, the total number of strings.
If there are no $0$s , the answer is easy, $1$. And if there is one $0$, the answer is equally easy, $8$.
We deal with two $0$s. So there will be six $1$s. Line them up in a row. The six $1$'s determine $7$ "gaps" (including the gaps at the two ends) that we can stuff a $0$ into. Choosing the $2$ gaps we use can be done in $\binom{7}{2}$ ways.
We deal next with three $0$s. So there will be five $1$'s, which determine $6$ "gaps" to stuff a $0$ into. Choosing the $3$ gaps we use can be done in $\binom{6}{3}$ ways.
Finally, we deal with four $0$s. We will have four $1$s, so $5$ gaps. The $4$ gaps we need for the $0$s can be chosen in $\binom{5}{4}$ ways.
That's all: if there are five or more $0$s, adjacency is unavoidable. Our count is therefore $\binom{9}{0}+\binom{8}{1}+\binom{7}{2}+\binom{6}{3}+\binom{5}{4}.$ (For fun, the easy cases of no $0$'s, one $0$ have been made to look like the others.)
Finally, calculate. We get $55$.