Solution:
Write
$$\begin{aligned} 3^{512}-1 & =\left(3^{256}+1\right)\left(3^{256}-1\right)=\left(3^{256}+1\right)\left(3^{128}+1\right)\left(3^{128}-1\right) \\ & =\cdots=\left(3^{256}+1\right)\left(3^{128}+1\right) \cdots(3+1)(3-1) \end{aligned}$$
Now each factor $3^{2^{k}}+1$, $k \geq 1$, is divisible by just one factor of $2$, since $3^{2^{k}}+1=\left(3^{2}\right)^{2^{k-1}}+1 \equiv 1^{2^{k-1}}+1=2 \pmod{4}$. Thus we get $8$ factors of $2$ here, and the remaining terms $(3+1)(3-1)=8$ give us $3$ more factors of $2$, for a total of $11$.