Disclaimer. The answer below is a variation of Bogomolov's argument which 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 ?

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 ?

Disclaimer. The answer below is a variation of Bogomolov's argument which 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 it true the Jacobian of $C$ is isogeneous to the square of an elliptic curve ?

Disclaimer. The answer below is a variation of Bogomolov's argument which 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 ?