I believe I managed to cook up an actual counterexample where both $C$ and $V$ are presheaf toposes. I'm going to leave my original attempt below since I still think it is instructive.
Let $V$ be presheaves over the category of finite sets $\mathsf{Fin}$ and let $C$ be the category of preasheaves over the poset $\mathbb{N}$ (i.e., towers of sets). Then $C$ carries a $V$ enrichment given as $C(X, Y)_A = C_0(X, m \mapsto Y_{m |A|})$ for $A \in \mathsf{Fin}$ and $X, Y \in C$. This makes $C$ into a (co)tensored $V$-category, but internal homs in $C$ are not enriched.
This does look quite criptic. Let me explain this as an instance of a general construction.
Let's take two small categories $P$ and $J$ with $P$ symmetric monoidal and acting on $J$. Then we set $V$ to be presheaves on $P$ with Day convolution and $C$ presheaves on $J$. We construct a $V$-enrichment as follows. Let $F_p \colon J \to J$ be the action of $p \in P$ on $J$. This extends via a left Kan extension to a functor $L_p \colon C \to C$ which has a right adjoint $R_p \colon C \to C$. We set $C(X, Y)_p = C_0(X, R_A Y)$. This makes $C(X, Y)$ into a $P$-presheaf and the composition operations are induced by the $P$-action of $J$. We also have (co)tensors. The tensor of $X \in C$ by a representable presheaf $P(-,p)$ is $L_p X$ and the contensor is $R_p X$ (and they extend by (co)limits to all presheaves).
$C$ is clearly cartesian closed and I believe there are many cases when the internal homs are not $V$-enriched. My old example below has a variant with $P = \mathbb{Z}/2$ (as a discrete monoidal category) and $J = \Delta$ with action by opposites.
In order to make $V$ cartesian, we can just take $P$ cartesian, e.g., $P = \mathsf{Fin}$. Then a natural candidate for $J$ is some category with finite coproducts so that $P$ acts by contensors. I took $J = \mathbb{N}$ to keep things easy, but I don't know if this is really the simplest choice. If we compute the functors $R_A$ for $A \in \mathsf{Fin}$, we indeed get $(R_A X)_m = X_{m |A|}$.
Now, take $A = \mathbf{2}$ be a two element set. Then $C(X \times Y, Z)_\mathbf{2} = C_0(X \times Y, R_\mathbf{2} Z) = C_0(X, (R_\mathbf{2} Z)^Y)$ and $C(X, Z^Y)_\mathbf{2} = C_0(X, R_\mathbf{2} (Z^Y))$ and $(R_\mathbf{2} Z)^Y$ and $R_\mathbf{2} (Z^Y)$ are clearly different in general.
Original answer
This is not an answer, but it is a bit too long for a comment. And perhaps it provides some hint towards an actual solution. Basically, I have a counterexample where the monoidal structure of $V$ is not cartesian, but I don't see if it can be modified to make it cartesian.
Let $V$ be the category of $\mathbb{Z}/2$-graded sets. The category $C = \mathsf{Cat}$ has a nice $V$-enrichment where even morphisms are covariant functors and odd morphisms are contravariant functors. It also has (co)tensors, the tensor of a category $X$ by a graded set $A$ is $X \times A_0 \sqcup X^{\mathrm{op}} \times A_1$ and the cotensor is $X^{A_0} \times (X^{\mathrm{op}})^{A_1}$. However, the internal homs in $C$ (with respect to the cartesian monoidal structure) are not enriched. Indeed, the odd elements of $C(X \times Y, Z)$ are functors that are contravariant in both $X$ and $Y$, but the odd elements of $C(X, Z^Y)$ are functors that are contravariant only in $X$.