In trying to understand better a certain weak choice axiom, I once concocted an example in sheaf theory which involves an exact pair in the sense of this question, without knowing it was a "thing" elsewhere.

In constructive mathematics, one choice principle which has been found useful is called the Presentation Axiom, Pax for short, which says roughly that any object $A$ (of one's given category) can be covered by an object $P$ which is projective: any further covering $Q \twoheadrightarrow P$ must split. You can imagine this applying for example to certain categories of sheaves of modules where one might want to consider projective resolutions. Or you can imagine it applying to certain categories of Heyting-valued sets or other toposes where the full axiom of choice is out of the question.

In constructive mathematics, it is also known as "CoSHEP" (Category of Sets Has Enough Projectives); a more philosophical justification (quoting the nLab) is

every set $A$ should have a ‘completely presented’ set of ‘canonical’ elements, that is elements given directly as they are constructed without regard for the equality relation imposed upon them. For canonical elements, equality is identity, so the BHK interpretation of logic justifies the axiom of choice for a completely presented set. This set is $P$, and $A$ is obtained from it as a quotient by the relation of equality on $A$. This argument can be made precise in many forms of type theory (including those of Martin-Löf and Thierry Coquand), which thus justify CoSHEP, much as they are widely known to justify dependent choice.

Now, Pax as stated above (for let's say a topos $E$) was actually the *external* version: an object $P$ is externally projective if the external hom-functor $\hom(P, -): E \to Set$ takes epimorphisms to epimorphisms. There's also an *internal* version of Pax which is often more appropriate: if $(-)^P: E \to E$ denotes the internal hom-functor, then we say $P$ is internally projective if $(-)^P$ preserves epimorphisms, and internal Pax would say that every object $A$ is covered by an internally projective $P$. What I had wanted was an example of a topos which validates external Pax but not internal Pax.

Presheaf toposes $E = Set^{C^{op}}$ (the category of functors $C^{op} \to Set$, where $C$ is a small category and $Set$ is just a vanilla category of sets in a ZFC background) always validate external Pax. It's easy to show that representable functors $\hom_C(-, a)$ are externally projective, as are coproducts of representable functors, and any object $F: C^{op} \to Set$ is covered by such a coproduct via a canonical map $\sum_{A \in Ob(C)} \sum_{x \in F A} hom_C(-, A) \twoheadrightarrow F$. So I wanted a $C$ which would force internal Pax to fail for $Set^{C^{op}}$.

The example I came up with is where $C$ is a generic exact pair, i.e., the poset $\mathbb{N} \cup \{a, b\}$ where the natural numbers $\mathbb{N}$ is ordered as usual and $a, b$ are incomparable elements that dominate every element of $\mathbb{N}$. We consider $C$ as a 'thin' category as is customary for posets. I claim that $A = \hom_C(-, a)$, or indeed any object $P$ covering $A$, cannot be internally projective. Specifically, that such $(-)^P: E \to E$ cannot preserve an evident epimorphism $F \twoheadrightarrow G$ where $F(n) = \{m \in \mathbb{N}: m \geq n\}$ and $F(a), F(b)$ are empty, where the restriction maps are given by the inclusions $F(n+1) \hookrightarrow F(n)$, and where $G$ is the presheaf with $G(n) = \ast$ (a one-point set) and $G(a), G(b)$ are empty.

To show the induced natural transformation $F^P \to G^P$ isn't an epimorphism, it suffices to show that the component $F^P(b) \to G^P(b)$ isn't a surjection between sets; in fact one may calculate that $F^P(b)$ is empty and $G^P(b)$ has exactly one element. Such calculations aren't exactly a spectator sport, so I'll wrap it up by making a few observations:

For general $F$ one may identify $F^P(b)$ with $Nat(\hom_C(-, b) \times P, F)$, where the right side indicates a set of natural transformations.

In the case where $P = \hom_C(-, a)$, we have in fact $\hom_C(-, b) \times \hom_C(-, a) \cong G$, and a natural transformation $G \to F$ boils down to having an element that belongs to the intersection of all the $F(n)$. Of course no such element exists since the intersection is empty.

So $F^{\hom_C(-, a)}(b)$ is empty, and $G^{\hom_C(-, a)}(b)$ has an element (in fact just one).

For general $P$ covering $\hom_C(-, a)$, there is a section $s: \hom_C(-, a) \to P$ (Yoneda lemma), and this section induces a map $F^P \to F^{\hom_C(-, a)}$, and thence a function $F^P(b) \to F^{\hom_C(-, a)}(b)$. The domain $F^P(b)$ has to be empty since the codomain is. On the other hand, $G^P(b)$ is nonempty because there is a map $\hom_C(-, b) \times P \to \hom_C(-, b) \times \hom_C(-, a) \cong G$.