We can always get non-principal ultrafilters on sets from non-discrete extremally disconnected spaces.
Let $X$ be a topological space. Let $\text{Clo}(X)$ denote the Boolean algebra of clopen subsets of $X$. If $X$ is extremally disconnected and zero dimensional, then $\text{Clo}(X)$ is a complete Boolean algebra, and for each non-isolated point $x_0\in X$, the set $\{R\in\text{Clo}(X):x_0\in R\}$ is a non-principal ultrafilter.
If $B$ is a complete Boolean algebra, and $U$ is an ultrafilter on $B$, then for each partition $p$ of $B$, we can define an ultrafilter $U_p=\{R\subseteq p:\bigvee R\in U\}$ on the set $p$.
Proposition: Suppose that $\kappa$ is a regular cardinal. Let $B$ be a $<\kappa$-complete Boolean algebra. Then an ultrafilter $U$ on $B$ is $<\kappa$-complete if and only if whenever $p$ is a partition of $B$ with $|p|<\kappa$, the ultrafilter $U_p$ is principal.
As a consequence, if $B$ is a complete Boolean algebra, and $U$ is a non-principal ultrafilter on $B$, then there is some partition $p$ where $U_p$ is a non-principal ultrafilter on $p$.
The above proposition is good for theory building, we can use a result with a simpler proof. We say that a subset $c$ of a lattice with zero $L$ is a cellular family if $a\wedge b=0$ whenever $a,b$ are distinct elements of $c$. If $A,B$ are subsets of a poset $P$, then write $A\preceq B$ and say that $A$ refines $B$ if for all $a\in A$ there exists a $b\in B$ with $a\leq b$.
In point-free topology, we have extremally disconnected frames, but in point-free topology, one does not need to worry about ultrafilters at all.
Added 2024-Sept-5
Ultraparacompactness lemma: If $B$ is a Boolean algebra and $R\subseteq B$ is a subset that has a least upper bound, then there exists a cellular family $C$ with $C\preceq R$ and $\bigvee C=\bigvee R$.
Proof: Order the collection of all cellular families $C$ with $C\preceq R$ by inclusion. Then by Zorn's lemma, there exists a maximal cellular family $C$ with respect to this ordering. Since $C\preceq R$, $\bigvee R$ is an upper bound of $C$. If $\bigvee R$ is not the least upper bound, then there is some upper bound $b$ of $C$ with $b<\bigvee R$. Let $a=(\bigvee R)\wedge b'$. Then $a=\bigvee_{r\in R}(r\wedge b')$, so there is some $r\in R$ with $0<r\wedge b'$, but $C\cup\{r\wedge b'\}$ is a cellular family with $C\cup\{r\wedge b'\}\preceq R$ which contradicts the maximality of $C$. Therefore, $\bigvee C=\bigvee R$. Q.E.D.
As a consequence of the ultraparacompactness, if $U$ is a non-principal ultrafilter on a complete Boolean algebra $B$, then there is some cellular family $p\subseteq B\setminus\{0\}$ with $p\subseteq B\setminus U$, but in this case $U_p$ is a non-principal ultrafilter on $p$.
Remarks (added by Gro-Tsen, 2024-08-20):
The above answer is correct, but I thought a few additional explanations would help illuminate it (any mistakes in what follows are entirely mine = Gro-Tsen's, of course).
The hypothesis that $X$ is zero-dimensional (on top of being extremally disconnected) is not a strong one: every regular extremally disconnected space is zero-dimensional, and some authors take this as part of the definition (e.g., in Hart, Nagata & Vaughan, Encyclopedia of General Topology, chapter g-01). Some kind of separation hypothesis is required: if an infinite set is endowed with the cofinite topology, it satisfies the property that the closure of an open set is still open, but it is not zero-dimensional and the construction outlined in the above answer fails. However, the latter space is not Hausdorff, something I had assumed in the question. There are Hausdorff (but not regular) extremally disconnected spaces which fail to be zero-dimensional (such as the “strong ultrafilter topology”, #113 in Steen & Seebach, Counterexamples in Topology), but this doesn't matter, because the proof works if we replace “zero-dimensional” by “totally separated”, where “totally separated” means the intersection of all clopen sets containing any $x_0 \in X$ is $\{x_0\}$, and this follows from extremally disconnected (Steen & Seebach, op. cit., fig. 9).
The infinite meet operation on $\operatorname{Clo}(X)$ is $\bigvee_{i \in I} R_i = \operatorname{closure}(\bigcup_{i\in I} R_i)$ as is easily checked (the “extremally disconnected” hypothesis ensures that this closure is open). So $p$ being a partition of $\operatorname{Clo}(X)$ means it's a set of clopen subsets of $X$ whose union is dense in $X$; and if $U$ is the ultrafilter $\{R\in\operatorname{Clo}(X):x_0\in R\}$ on $\operatorname{Clo}(X)$. the ultrafilter $U_p$ will consist of subsets of $p$ such that $x_0$ is in the closure of the union. (I'm spelling this out because I had initially read this as $x_0$ just being in the union and couldn't figure out how this could fail to be principal.)
To say that a ultrafilter $U$ on a complete (resp. ${<}\kappa$-complete) Boolean algebra $B$ is complete (resp. ${<}\kappa$-complete) means that for any $I$ (resp. one with $|I| < \kappa$), if $R_i \in U$ for all $i\in I$ then $\bigwedge_{i\in I} R_i \in U$ (or, of course, dually: if $\bigvee_{i\in I} R_i \in U$ then $R_i \in U$ for some $i\in I$). The crucial fact used in the proof is now that $\operatorname{Clo}(X)$ is complete (as pointed out in the previous bullet), but $\{R\in\operatorname{Clo}(X):x_0\in R\}$ is not complete unless $x_0$ is isolated (because, the space being zero-dimensional, the intersection of all clopen neighborhoods of $x_0$ is $\{x_0\}$ as $X$ is totally separated; and unless $x_0$ is isolated this means that their meet in $\operatorname{Clo}(X)$ is $\varnothing$).
Here is a proof of the stated proposition. One direction is trivial: assume $B$ is a ${<}\kappa$-complete Boolean algebra and $U$ is a ${<}\kappa$-complete ultrafilter on $B$; if $p$ is a partition with $|p|<\kappa$ then $\bigvee_{R\in p} R = \top$ so by ${<}\kappa$-completeness of $U$ there is $R\in p$ such that $R\in U$ which means $U_p$ is principal. Now for the other direction:
Assume $B$ is a ${<}\kappa$-complete Boolean algebra and $U$ is an ultrafilter on $B$ which is not $\lambda$-complete for some $\lambda<\kappa$. Using the well-ordering principle, this means there are $R_\xi$ for $\xi<\lambda$ (identifying $\lambda$ with the smallest ordinal of that cardinal) such that $R_\xi \in U$ but $\bigwedge_{\xi<\lambda} R_\xi \not\in U$.
Now define the elements $P_\xi := (\neg R_\xi) \wedge \bigwedge_{\iota<\xi} R_\iota$ for $\xi<\lambda$, where $\neg R_\xi$ denotes the Boolean complement of $R_\xi$ (starting with $P_0 = \neg R_0$), and also $P_\lambda := \bigwedge_{\xi<\lambda} R_\xi$. Clearly, $P_\iota\wedge P_\xi = \bot$ if $\iota<\xi\leq\lambda$ (because $P_\iota\leq\neg R_\iota$ and $P_\xi\leq R_\iota$).
By transfinite induction on $\xi\leq\lambda$ we show that
$$
\Big(\bigvee_{\iota<\xi} P_\iota\Big) \vee \Big(\bigwedge_{\iota<\xi} R_\iota\Big) = \top
$$
For $\xi=0$ this is trivial. At successor $\xi$ this follows from the induction hypothesis by writing
$$
\bigwedge_{\iota<\xi} R_\iota = ((\neg R_\xi) \vee R_\xi) \wedge \big(\bigwedge_{\iota<\xi} R_\iota\big) = P_\xi\, \vee \, \big(\bigwedge_{\iota<\xi+1} R_\iota\big)
$$
And at limit $\xi$ it follows by using the distributivity of finite $\vee$ on infinite $\bigwedge$ (see e.g., here) to write
$$
\Big(\bigvee_{\iota<\xi} P_\iota\Big) \vee \Big(\bigwedge_{\iota<\xi} R_\iota\Big) = \bigwedge_{\iota<\xi} \Bigg(\big(\bigvee_{\iota<\xi} P_\iota\big) \vee R_\iota\Bigg)
$$
and each $\big(\bigvee_{\iota<\xi} P_\iota\big) \vee R_\iota$ is $\geq \big(\bigvee_{\gamma\leq\iota} P_\gamma\big) \vee \big(\bigwedge_{\gamma\leq\iota}R_\gamma\big)$ which is $\top$ by the induction hypothesis. This concludes the induction, and in particular we have $\big(\bigvee_{\xi<\lambda} P_\xi\big) \vee P_\lambda = \top$, in other words the $P_\xi$ for $\xi\leq\lambda$ form a partition of $B$.
But none of the $P_\xi$ is in $U$: for $\xi<\lambda$ this is because $\neg R_\xi \not\in U$; and for $P_\lambda$ this is because $\bigwedge_{\xi<\lambda} R_\xi \not\in U$. So the ultrafilter $U_p$ on $p = \{ P_\xi : \xi\leq\lambda \}$ is not principal. This concludes the proof of the proposition. ∎
