Let $k$ be an infinite cardinal. We already know there are exactly $2^{2^k}$ distinct non-principal ultrafilter on $k$.
Here The set of ultrafilters on an infinite set is uncountable. And proof uses that independent family of cardinality $2^k$.
"Can we force something upon that independent family such that the ultrafilters generated have character $2^k$?"
But my question for the post is: Are there exactly $2^{2^k}$ distinct non-principal ultrafilters with character equal to $2^k$?
Definition: Character of ultrafilter $\mathcal{U}$ is the least cardinality of a filter base of $\mathcal{U}$.
Let $I$ be an index set of cardinality $|I|=2^\kappa$. By Hausdorff's theorem proved in the linked post, there is a collection $\{A_i\}_{i\in I}\subseteq\mathcal{P}(\kappa)$ such that $A_i\ne A_j$ for $i\ne j$ and the $A_i$ generate a free Boolean sub-algebra of $\mathcal{P}(\kappa)$.
Notation: If $S\subseteq\kappa$, $S^\complement:=\kappa\setminus S$.
Let $\mathfrak T=\{T\subseteq I\colon |T|=\aleph_0\}$. For $T\in\mathfrak T$ define $\mathcal S_T:=\left(\bigcup_{i\in T}A_i^\complement\right)\in\mathcal{P}(\kappa)$.
Claim: The collection $\mathfrak L:=\{A_i\colon i\in I\}\cup\{\mathcal S_T\colon T\in\mathfrak T\}$ has the finite intersection property.
Proof: Let $F$ be a finite subset of $I$. Let $J$ be a finite subset of $\mathfrak T$. For each $T\in J$, since $|T|=\aleph_0$ and $F$ is finite, there is an index $i_T\in T\setminus F$. By definition of $\mathcal S_T$, $A_{i_T}^\complement\subseteq \mathcal{S}_T$. Therefore $$ \left(\bigcap_{f\in F} A_f\right)\cap\left(\bigcap_{T\in J}\mathcal{S}_T\right)\supseteq\left(\bigcap_{f\in F} A_f\right)\cap\left(\bigcap_{T\in J}A_{i_T}^\complement\right)\ne\emptyset\tag{1} $$ the right hand side being non-empty because $\{A_i\}_{i\in I}$ generate a free Boolean algebra. Note that there may be repetitions among the $(i_T)_{T\in J}$, but this is ok, as long as all the $i_T\not\in F$ - which is how they were chosen. This completes proof of the claim.
Now let $\mathfrak u$ be an ultrafilter such that $\mathfrak L\subseteq\mathfrak u$. Let $\mathcal B\subseteq \mathfrak u$ be a filter-base. Since $\mathfrak u\subseteq\mathcal{P}(\kappa)$, it's clear that $|\mathcal B|\le2^\kappa$. We show that $|\mathcal B|\ge 2^\kappa$.
For $B\in\mathcal B$ let $I_B:=\{i\in I\colon B\subseteq A_i\}$. We show that $I_B$ is finite. Assume conversely that $I_B$ is infinite and let $T\subseteq I_B$ such that $|T|=\aleph_0$. Then $B\subseteq A_i$ for all $i\in T$. Therefore $B\subseteq \bigcap_{i\in T} A_i$. Therefore $\bigcap_{i\in T} A_i\in\mathfrak u$. But $\mathcal S_T=\left(\bigcap_{i\in T} A_i\right)^\complement$ and $S_T\in\mathfrak L\subseteq\mathfrak u$. This is a contradiction to $\mathfrak u$ being a filter. Hence $I_B$ is finite.
But now, since $\mathcal B$ is a filter-base, for each $i\in I$ there has to be some $B\in\mathcal B$ such that $B\subseteq A_i$. Since each $I_B$ is finite, and $|I|=2^\kappa$. It follows that $|\mathcal B|\ge 2^\kappa$. This concludes the proof.
We showed how to get one ultrafilter with character $2^\kappa$. Note however, that in the family $\{A_i\}_{i\in I}$ we started with, we can complement each member independently and still remain with a family that generates a free Boolean algebra. Moreover, each ultrafilter obtained by choosing such a combination of $A_i$ or $A_i^\complement$ will be distinct. Hence there are $2^{2^\kappa}$ distinct ultrafilters with character $2^\kappa$.