8. The sequence $\left\{a_{n}\right\}$ is defined as follows:
$$
a_{0}=0, a_{1}=1, a_{n}=2 a_{n-1}+a_{n-2}(n>1) \text {. }
$$

Then the number of $n$ that satisfies $2^{2011} \mid a_{n}$ is $\qquad$ .
8. Infinitely many.

From the characteristic equation $x^{2}-2 x-1=0$, its two roots are $1+\sqrt{2}, 1-\sqrt{2}$.
From the initial conditions, the general term of the sequence is
$$
a_{n}=\frac{(1+\sqrt{2})^{n}-(1-\sqrt{2})^{n}}{2 \sqrt{2}} \text {. }
$$

By the binomial theorem, we have
$$
\begin{aligned}
& (1+\sqrt{2})^{2 n}-(1-\sqrt{2})^{2 n} \\
= & {\left[(1+\sqrt{2})^{n}+(1-\sqrt{2})^{n}\right] . } \\
& {\left[(1+\sqrt{2})^{n}-(1-\sqrt{2})^{n}\right] } \\
= & 2\left(C_{n}^{0}+C_{n}^{2} \cdot 2+\cdots\right)\left[(1+\sqrt{2})^{n}-(1-\sqrt{2})^{n}\right] .
\end{aligned}
$$

Thus, if $2^{k} \mid a_{n}$, then $2^{k+1} \mid a_{2 n}$.
Since $a_{2}=2$, by mathematical induction, we can prove:
If $2^{k} \mid n$, then $2^{k} \mid a_{n}$.
Therefore, there are infinitely many $n$ that satisfy the requirement.