Assume you have a proper filter $F$ that avoids both $b$ and $\neg b$. Then, you could consider the filter generated by $F\cup\{b\}$ - which is to say the smallest filter $F'$ containing $F$ and $b$.

Since $F$ was a proper filter it follows that $0\not\in F$. If $0\in F'$, then this means that there is some $f\in F$ such that $b\wedge f = 0$. Now, $\neg b=0\vee\neg b=(b\wedge f)\vee\neg b=(b\vee\neg b)\wedge(f\vee\neg b)=1\wedge(f\vee\neg b)=f\vee\neg b$. Thus $f≤\neg b$, which means that $\neg b\in F$, which is a contradiction.

Hence, $0\not\in F'$, and thus $F'$ is a proper ideal strictly containing $F$.

Now use Zorn's lemma, as Robin Chapman suggested, to climb up to a maximal filter. The argument above, though, shows that a maximal filter cannot avoid a pair $b, \neg b$.

Assume you have a proper filter $F$ that avoids both $b$ and $\neg b$. Then, you could consider the filter generated by $F\cup\{b\}$ - which is to say the smallest filter $F'$ containing $F$ and $b$.

Since $F$ was a proper filter it follows that $0\not\in F$. 

If $0\in F'$, then this means that there is some $f\in F'$ such that $b\wedge f = 0$. Now, $\neg b=0\vee\neg b=(b\wedge f)\vee\neg b=(b\vee\neg b)\wedge(f\vee\neg b)=1\wedge(f\vee\neg b)=f\vee\neg b$. Thus $f≤\neg b$, which means that $\neg b\in F'$.

Since $\neg b\in F'$, either $\neg b\in F$ or $\neg b$ may be acquired by meets and upwards closures from $F\cup\{b\}$. Say $b\wedge f≤\neg b$ for some $f\in F$. Then $b\wedge f= b\wedge f\wedge\neg b = b\wedge\neg b\wedge f = 0\wedge f = 0$ for an $f\in F$ and by the above argument, we derive $\neg b\in F$. This is a contradiction, from which we can derive that $0\not\in F$.

Hence, $0\not\in F'$, and thus $F'$ is a proper ideal strictly containing $F$.