**Disclaimer.** The answer below is a variation of Bogomolov's argument and it would not come to be without Dmitri's answer. If you feel like upvoting this, please upvote his answer.

**Curves on products of isogeneous elliptic curves.**
As already suggested in the body of the question, if we start with a pair of elliptic
curves, say $E_1$ and $E_2$, admitting a non-constant morphism $f : E_1 \to E_2$ then
given any point $p \in X=E_1 \times E_2$ we have infinitely many elliptic curves with self-intersection zero on $X$ passing through $p$. It suffices to consider translates of the graphs of endomorphisms of $E_2$ (there are at least $\mathbb Z$ of them) composed with $f$.

If we blow-up $p$ then we get a surface $S$ containing infinitely many (elliptic) curves
with negative self-intersection.

**Jacobians of genus $2$ curves.** As pointed out in Dmitri's answer the
natural morphism
$$
\mathrm{Sym}^2 C \to \rm{Pic}^2(C) \cong \rm{Jac}(C)
$$
identifies $\mathrm{Sym}^2 C$ with the blow-up of $\rm{Jac}(C)$ at a point.
Thus if
we have a genus $2$ curve with Jacobian isogenous to the square of an elliptic curve
then the discussion in the previous paragraph shows that $C^2$ has infinitely many
curves of negative self-intersection since we can pull-back the
negative curves on $\mathrm{Sym}^2 C$ through the natural morphism $C^2 \to \rm{Sym}^2 C$.
Notice also that the negative curves have unbounded intersection with the diagonal $\Delta \subset C^2$. It is not hard to verify that the pull-backs of the negative elliptic curves to $C^2$ will have unbounded genus.

**Explicit example.**
If $C$ is a genus $2$ curve admitting a morphism
$\pi : C \to E$ to an elliptic curve $E$ then $\rm{Jac}(C)$ is isogeneous to the product of $E$ with another
elliptic curve $E'$ ( the connected component of the kernel of $\pi$ through zero).
Automorphisms of $C$ act naturally on $\rm{Jac}(C)$. If there is an element

$\varphi \in \mathrm{Aut}(C)$ with induced action on $\rm{Jac}(C)$ not preserving $E'$ then $E$ is isogeneous to $E'$ since $$\pi_* \circ \varphi_* : \rm{Jac}(C) \to \rm{Jac}(E)\cong E$$ restricted to $E'$ is an isogeny. Therefore $\rm{Jac}(C)$ is isogeneous to the square of $E$.

To have a concrete example we can take $C = \lbrace y^2 = x^6 - 1\rbrace$ which maps to $E =\lbrace y^2 = x^3 -1\rbrace$ and has automorphism group isomorphic to $\mathbb Z_3 \rtimes D_8$ (which is not the automorphism group of any elliptic curve). From the discussion above
it follows that $C^2$ has infinitely many curves of negative self-intersection and unbounded genus.

**Question.** Suppose $C$ is genus $2$ curve such that $C^2$ contains infinitely many
curves of negative self-intersection. Is the Jacobian of $C$ isogeneous
to the square of an elliptic curve ?