 Solution

Yes. Order the proper filters containing the cofinite filter by inclusion. The union of every chain is again a proper filter containing $\mathcal{F}$: any finite collection of its members already lies in one filter in the chain. Zorn lemma gives a maximal proper filter $\mathcal{U}$.

A maximal proper filter is an ultrafilter. Indeed, if $A\notin \mathcal{U}$, adjoining $A$ cannot generate another proper filter, so some $U\in \mathcal{U}$ has $U\cap A=\varnothing$. Thus $U\subseteq A^c$, and upward closure gives $A^c\in \mathcal{U}$. Finally, $\mathcal{U}$ cannot equal any principal ultrafilter ${\mathcal{U}}_n$, because it contains the cofinite set $\mathbb{N}\setminus \{n\}$ whereas ${\mathcal{U}}_n$ does not. Thus $\mathcal{U}$ is a nonprincipal ultrafilter on $\mathbb{N}$.