The bounded distributive lattices which every prime filter is principal can be characterized in terms of Priestley spaces, Priestley duality and certain posets (these are some results that I proved a while ago but I would not be surprised if someone else has already obtained these results already).

Let me review some facts about Priestley duality.

A partially ordered compact space $X$ is said to be a Priestley space if whenever $x\not\leq y$ then there is a clopen upper set $U$ with $x\in U$ and $y\not\in U$. If $X$ is a Priestley space, then let $\mathcal{C}_{d}(X)$ be the collection of all clopen downwards closed subsets of $X$. Then $\mathcal{C}_{d}(X)$.

Suppose that $D$ is a bounded distributive lattice. Let $\mathcal{I}_{p}(D)$ be the set of all prime ideals in $D$ ordered by inclusion. If $a\in D$, then let $C_{a}=\{I\in\mathcal{I}_{p}(D)|a\not\in I\}$. Then $\{C_{a}\setminus C_{b}|a,b\in D\}$ is a basis for a Priestley space topology on $\mathcal{I}_{p}(D)$.

Priestley duality states that if $D$ is a distributive lattice, then $D\simeq \mathcal{C}_{d}(\mathcal{I}_{p}(D))$ and if $X$ is a priestley space, then $X\simeq\mathcal{I}_{p}(\mathcal{C}_{p}(X))$ and that these isomorphisms are natural.

Suppose that $(X,\leq,\mathcal{T})$ be an ordered topological space. Then we shall call $(X,\leq,\mathcal{T})$ an upper limit system if each set of the form $\downarrow x$ is closed and $\{\downarrow x\cap U|x\in X,\text{$U$ is an open upwards closed set}\}$ forms a basis for the topology $\mathcal{T}$.

Take note that if $(X,\leq,\mathcal{T})$ is an upper limit system, then each set $\downarrow x$ is clopen and each lower set is open.

$\mathbf{Lemma}$ Suppose that $X$ is a Priestley space. Let $x\in X$. Then

The ideal $\{C\in\mathcal{C}_{d}(X)|x\not\in C\}$ is principal in $\mathcal{C}_{d}(X)$ if and only if $\uparrow x$ is clopen.

The filter $\{C\in\mathcal{C}_{d}(X)|x\in C\}$ is principal in $\mathcal{C}_{d}(X)$ if and only if the set $\downarrow x$ is clopen.

$\mathbf{Proof}$

We shall only prove part 2 since part 1 is identical.

$\leftarrow$ If $\downarrow x$ is clopen, then clearly $\downarrow x$ is the least element in $\{C\in\mathcal{C}_{d}(X)|x\in C\}$, so $\{C\in\mathcal{C}_{d}(X)|x\in C\}$ is principal in $\mathcal{C}_{d}(X)$.

$\rightarrow$ Suppose now that $\{C\in\mathcal{C}_{d}(X)|x\in C\}$ is principal. Then let $R$ be the least element in $\{C\in\mathcal{C}_{d}(X)|x\in C\}$. Then clearly $\downarrow x\subseteq R$. Suppose now that $\downarrow x$ is a proper subset of $R$. Then let $y\in R\setminus\downarrow x$. Then since $y\not\leq x$, there is a clopen downwards closed subset $S$ with $x\in S$ but $y\not\in S$. Therefore $S\in\{C\in\mathcal{C}_{d}(X)|x\in C\}$ but $R\not\subseteq S$ contradicting the minimality of $R$. We therefore conclude that $\downarrow x=R$ after all. Therefore $\downarrow x$ is clopen.

$\mathbf{QED}$

The following proposition is not too hard to prove.

$\mathbf{Proposition}$ Let $X$ be a Priestley space. Then the following are equivalent.

$X$ is an upper limit space.

Every set of the form $\downarrow x$ is clopen.

Every filter in $\mathcal{C}_{d}(X)$ is principal.

Suppose that $X$ is a poset. Then the minimal upper limit topology on $X$ is the topology generated by the subbasis $\{\downarrow x|x\in X\}\cup\{(\downarrow x)^{c}|x\in X\}$. The minimal upper limit topology is clearly generated by the basis consisting of sets of the form $\downarrow x\cap(\downarrow y_{1})^{c}\cap...\cap(\downarrow y_{n})^{c}$.

Clearly every poset whose minimal upper limit topology is compact is a Priestley space in the minimal upper limit topology and every Priestley space which is an upper limit topology is the minimal upper limit topology. We conclude that by Priestley duality the bounded distributive lattices where every filter is principal are in a one-to-one correspondence with the posets which are compact in the minimal upper limit topology. The following theorem characterizes the posets which are compact in the minimal upper limit topology.

$\mathbf{Theorem}$ Suppose that $X$ is a poset. Then $X$ is compact in the minimal upper limit topology if and only if

for each $B\subseteq X$ there are lower bounds $a_{1},...,a_{n}$ of $B$ such that if $a$ is a lower bound of $B$, then $a\leq a_{i}$ for some $i\in\{1,...,n\}$.

for each $B\subseteq X$ there are $b_{1},...,b_{n}\in B$ where if $a$ is a lower bound of $\{b_{1},...,b_{n}\}$ then $a$ is a lower bound of $B$.

$\mathbf{Proof}$

$\rightarrow$ Suppose that $X$ is compact in the minimal upper limit topology. Suppose that $B\subseteq X$ is a subset.

Then $\downarrow B=\bigcap_{b\in B}\downarrow b$ is closed being the intersection of closed sets. However, $\downarrow B$ is a downwards closed set, so $\downarrow B$ is also an open set. Thus, $\downarrow B$ is clopen.

Let's now show that statement 1 holds. Since $\downarrow B$ is open and $\{\downarrow x|x\in\downarrow B\}$ is an open cover of $X$, there are $a_{1},...,a_{n}\in B$ so that $\downarrow B=\downarrow a_{1}\cup...\cup\downarrow a_{n}$. Therefore, $a_{1},...,a_{n}$ are lower bounds of $B$ so that if $a$ is a lower bound of $B$ then $a\leq a_{i}$ for some $i\in\{1,...,n\}$.

We shall now show statement 2 holds.
We have $(\downarrow B)^{c}=\bigcup_{b\in B}(\downarrow b)^{c}$. Therefore, since $(\downarrow B)^{c}$ is compact, there are $b_{1},...,b_{n}$ with $(\downarrow B)^{c}=(\downarrow b_{1})^{c}\cup...\cup(\downarrow b_{n})^{c}$, so we have $\downarrow B=\downarrow b_{1}\cap...\cap \downarrow b_{n}$. Therefore, $b_{1},...,b_{n}$ are elements in $B$ such that if $a$ is a lower bound of $\{b_{1},...,b_{n}\}$ then $a$ is also a lower bound of $B$.

$\leftarrow$ To prove this direction, we shall use Alexander's subbase theorem. In particular, we shall show that $X$ is compact by showing that every cover from the basis $\{\downarrow x|x\in X\}\cup\{(\downarrow x)^{c}|x\in X\}$ has a finite subcover.

Suppose that $A,B\subseteq X$ are subsets such that
$\mathcal{U}=\{\downarrow a|a\in A\}\cup\{(\downarrow b)^{c}|b\in B\}$ covers $X$.

Let $\mathcal{V}=\{(\downarrow b)^{c}|b\in B\}\cup\{\downarrow x|x\in\downarrow B\}$. Then $\mathcal{V}$ covers $X$. I claim that $\mathcal{V}$ refines $\mathcal{U}$. Suppose that $V\in\mathcal{V}$. If $V=(\downarrow b)^{c}$, then $V\subseteq V\in\mathcal{U}$. If $V=\downarrow x$ for some $x\in\downarrow B$, then $x\in U$ for some $U\in\mathcal{V}$. However, since $x\in\downarrow B$, if $b\in B$, then $x\in\downarrow b$, so $x\not\in(\downarrow b)^{c}$. Therefore $U\neq(\downarrow b)^{c}$ for $b\in B$. Therefore, $U=\downarrow a$ for some $a\in A$. We conclude that $V=\downarrow x\subseteq\downarrow a=U$. We conclude that $\mathcal{V}$ refines $\mathcal{U}$.

We shall now show that $\mathcal{V}$ has a finite subcover.

By 2, there are $b_{1},...,b_{n}\in B$ where $a$ is a lower bound of $b_{1},...,b_{n}$ if and only if $a$ is a lower bound of $B$. In other words, we have $\downarrow b_{1}\cap...\cap\downarrow b_{n}=\downarrow B$. Therefore, we have $(\downarrow b_{1})^{c}\cup...\cup(\downarrow b_{n})^{c}=\bigcup_{b\in B}(\downarrow b)^{c}=(\downarrow B)^{c}$.

Similarly, there are lower bounds $a_{1},...,a_{m}$ of $B$ so that if $a$ is a lower bound of $B$, then $a\leq a_{i}$ for some $i\in\{1,...,m\}$. Therefore, we have $\downarrow a_{1}\cup...\cup\downarrow a_{m}=\downarrow B$. Therefore, we have
$X=\downarrow a_{1}\cup...\cup\downarrow a_{m}\cup(\downarrow b_{1})^{c}\cup...\cup(\downarrow b_{n})^{c}$. Thus
$\{\downarrow a_{1},...,\downarrow a_{m},(\downarrow b_{1})^{c},...,(\downarrow b_{n})^{c}\}$ is a finite subcover of $\mathcal{V}$. Since $\mathcal{V}$ refines $\mathcal{U}$, the cover $\mathcal{U}$ has a finite subcover as well. Therefore, by Alexander's subbase theorem, the space $X$ is compact. $\mathbf{QED}$

Therefore we conclude that the bounded distributive lattices where every prime filter is principal are in a one-to-one correspondence with the posets satisfying the conditions of the above theorem. We conclude with a proof of a proposition that Dominic van der Zypen has claimed in his answer.

$\mathbf{Proposition}$ Every infinite bounded distributive lattice has a non-principal prime filter or a non-principal prime ideal.

$\mathbf{Proof}$ Suppose that $X$ is the Priestley space for a bounded distributive lattice where every ideal and filter is principal. Then each set of the form $\downarrow x$ and $\uparrow x$ is clopen, so $\{x\}=\downarrow x\cap\uparrow x$ is clopen. Therefore, since every point in $X$ is isolated, we conclude that $X$ is finite, so its corresponding bounded distributive lattice is also finite.