I believe the answer is yes: via the ultrafilter functor, $Set$ is comonadic over $Stone$ (and also over $CpctHaus$). Instead of the plain comonadicity theorem, let's use the crude comonadicity theorem, which says the following:

Let $F: C \to D$ be a conservative left adjoint such that $C$ has and $F$ preserves equalizers of reflexive pairs. Then $F$ is comonadic.

So consider a reflexive pair in $Set$, i.e. we have $f,g: A \rightrightarrows B$ which are both sections of a map $p: B \to A$. Let $E = \{a \in A \mid f(a) = g(a)\}$ be the equalizer. Let $\mathcal F$ be an ultrafilter on $A$ such that $f_\ast(\mathcal F) = g_\ast(\mathcal F)$, i.e.

$(\ast)$ For all $V \subseteq B$ we have $f^{-1}(V) \in \mathcal F \Leftrightarrow g^{-1}(V) \in \mathcal F$.

It will suffice to show that $E \in \mathcal F$. Suppose not. Then $(A \setminus E) \in \mathcal F$. Let $V \subseteq B$ be the set $V = \{f(a) \mid a \in (A \setminus E)\}$. Then we have $f^{-1}(V) = (A \setminus E) \in \mathcal F$, but $g^{-1}(V) = \emptyset \not \in \mathcal F$, contradicting $(\ast)$. Thus we must have $E \in \mathcal F$ after all, and so the ultrafilter functor preserves this reflexive equalizer.

Note that the same proof shows that the ultrafilter functor $\beta: Set \to \mathcal C$ preserves reflexive equalizers for any of $\mathcal C \in \{ Stone, CpctHaus, Set\}$. In particular, the ultrafilter functor exhibits $Set$ as comonadic over the category of compact Hausdorff spaces as well as over the category of Stone spaces.

I believe the answer is yes: via the ultrafilter functor, $Set$ is comonadic over $Stone$ (and also over $CpctHaus$). Instead of the plain comonadicity theorem, let's use the crude comonadicity theorem, which says the following:

Let $F: C \to D$ be a conservative left adjoint such that $C$ has and $F$ preserves equalizers of coreflexive pairs. Then $F$ is comonadic.

So consider a coreflexive pair in $Set$, i.e. we have $f,g: A \rightrightarrows B$ which are both sections of a map $p: B \to A$. Let $E = \{a \in A \mid f(a) = g(a)\}$ be the equalizer. Let $\mathcal F$ be an ultrafilter on $A$ such that $f_\ast(\mathcal F) = g_\ast(\mathcal F)$, i.e.

$(\ast)$ For all $V \subseteq B$ we have $f^{-1}(V) \in \mathcal F \Leftrightarrow g^{-1}(V) \in \mathcal F$.

It will suffice to show that $E \in \mathcal F$. Suppose not. Then $(A \setminus E) \in \mathcal F$. Let $V \subseteq B$ be the set $V = \{f(a) \mid a \in (A \setminus E)\}$. Then we have $f^{-1}(V) = (A \setminus E) \in \mathcal F$, but $g^{-1}(V) = \emptyset \not \in \mathcal F$, contradicting $(\ast)$. Thus we must have $E \in \mathcal F$ after all, and so the ultrafilter functor preserves this coreflexive equalizer.

Note that the same proof shows that the ultrafilter functor $\beta: Set \to \mathcal C$ preserves coreflexive equalizers for any of $\mathcal C \in \{ Stone, CpctHaus, Set\}$. In particular, the ultrafilter functor exhibits $Set$ as comonadic over the category of compact Hausdorff spaces as well as over the category of Stone spaces.