Left Kan extension along Yoneda of pullback-preserving functor preserving pullbacks
If you are wiling to assume that $C$ has a terminal object $1 \in C$, which I assume is the case as you said all finite products, you can do the following:
(As it is not clear if you are interested in a $1$-categorical statement or an $\infty$-categorical statement I'll write the proof in a very formal style which should accomodate $\infty$-categories as well with a little bit of work)
Saying that $F:C \rightarrow D$ preserve fiber products means that $F_{/1}: C \rightarrow D_{/F(1)}$ preserves all finite limits (as it preserve pullback and the initial objects).
In particular the induced functor $Prsh(C) \rightarrow Prsh(D_{/F(1)}) $ preserves all finite limits by your "well known fact".
But $Prsh(D_{/F(1)})$ identifies canonically with $Prsh(D)_{/y_D(F(1))}$ and $y_D(F(1))$ is $F_!(y_C(1))$ with $y_C$ being the terminal object of $Prsh(C)$.
In the end the functor: $F_!$ preserve all finite limits when seen as a functor from $Prsh(C)$ to $Prsh(D)_{/F_!(1)}$ but this means that $F_!$ preserve fiber product as asked.
I have ignored the fact that $D$ does not have all finite limits as, as far as I know, this has never been part of the assumption of your "well known fact".
Indeed, if $C$ has all finite limits and $F:C \rightarrow D$ preserve them, then in order to check that $F_!$ preserve finite limits one justs need to check that for all $d \in D$ the functor $X \mapsto Hom(d,F_! ( X))$ from $Prsh(C)$ to $Set$ preserves all finite limits.
This functor is the left Kan extention along the Yoneda embedings of $C$ of the functor $C \rightarrow Set$ which send $c$ to $Hom(d,F(c))$, which commutes to finite limits by definition.
So in the end the relevant fact is that if $C$ has finite limits $F:C \rightarrow Set$ preserves them then its left Kan extention $Prsh(C) \rightarrow Set$ also preserve finite limits. But existence of limits in $D$ never really played a role.
I'm adding a counter example to show that the assumption that $C$ has a terminal object cannot be completely relaxed.
Let $C$ be a group $G$, seen as a one object category. And let $D$ be the terminal category. With $F:C \rightarrow D$ the unique functor.
The category $C$ have no terminal object but it has all fibered product (because all map are isomorphisms), and the functor $F$ preserve all fibered product limits (in fact we have both that any functor out of $C$ and that any functor to $D$ preserves all fibered product)
$F_!$ is the functor from $G$-set to set which send any $G$-set to its quotient $X/G$.
$F_!$ send the terminal object to the terminal object but does not preserve product: if $G$ denotes $G$ with its multiplication action on itself then $F_! (G \times G) = G$ but $F_!(G)=\{*\}$.
So $F_!$ does not preserve fibered product.
Two remarks:
I do not know if there are other condition than having a terminal object that allow this sort of result to work.
This examples suggest that things might work better if one work with $\infty$-categories and $Prsh$ means space valued presheaves. I do not know the status of this (Edit: see me second answer about this specific point).
My previous answer left open the following:
Proposition: Let $C$ be a small $\infty$-category with all fiber products, let $\mathcal{T}$ be an $\infty$-topos and let $F : C \rightarrow \mathcal{T}$ be a functor preserving fiber products. Then the left Kan extention: $$ \widehat{F} : \text{Prsh}(C) \rightarrow \mathcal{T} $$ also preserves fiber product.
Indeed in the previous answer I showed that the statement was true as soon as $C$ has a terminal object and I have given an example showing that the $1$-categorical version of the proposition is false. But the example I have given somehow suggested that the proposition might be true in the $\infty$-categorical settings. I recently needed an answer to this question so I thought about it, and I think I have a proof of the proposition above.
Proof: As $C$ has all fiber product, and $F$ preserves them, one has that for each object $c\in C$, the slice category $C/c$ has all finite limits, and the functor induced by $F$:
$$ F/c : C/c \rightarrow \mathcal{T}/F(c) $$
preserves them. In particular the left Kan extention:
$$ \widehat{F/c} : \text{Prsh}(C/c) \rightarrow \mathcal{T}/F(c) $$
preserve all finite limit by the results of Lurie. This functor is isomorphic to:
$$ \widehat{F}/c : \text{Prsh}(C)/c \rightarrow \mathcal{T}/F(c) $$
It follows in particular that $\widehat{F}$ preserves all fiber product of the form $X \times_c Y$ where $c$ is a representable object. Note that up to this point, everything also works in the $1$-categorical case, and this is exactly how we concluded in the special case where $C$ has a terminal object. The idea is now to use the descent property to extend this to more general pullback.
Let $V \rightarrow X$ be any morphism in $\text{Prsh}(C)$, and write $X$ as the canonical colimits:
$$ X = \underset{c \in Elt(X)}{\text{colim }} c $$
(where $Elt(X)$ is the category of elements of $X$ and I have identified objects of $C$ with their image by the Yoneda embedding)
For each elements $c \rightarrow X$, let $V_c$ be the fiber of $V \rightarrow X$ over $c$.
By descent, one has:
$$ V = \underset{c \in Elt(X)}{\text{colim }} V_c $$
Now, as $\widehat{F}$ preserves all pullbacks whose bottom corner is representable, it follows that the the natural transforation: $\widehat{F}(V_c) \rightarrow F(c)$ is cartesian (in $c$), and hence, by descent in $\mathcal{T}$ it follows that all the maps $\widehat{F}(V_c) \rightarrow F(c)$ are pullback of the maps between the colimits:
$$ \widehat{F}(V) \simeq \underset{c \in Elt(X)}{\text{colim }} \widehat{F}(V_c) \rightarrow \underset{c \in Elt(X)}{\text{colim }} F(c) \simeq \widehat{F}(X)$$
Which shows that $\widehat{F}$ also preserves all pullback of the form $V \times_X c$ as soon as $c$ is representable.
To conclude, one either use a again a similar (but simpler) colimit/descent argument in the variable $c$, or one simply write (using descent) that the functor $\text{Prsh}(C)/X \rightarrow \mathcal{T}/\widehat{F}(X)$ can be decomposed as a limits:
$$ \text{Prsh}(C)/X \simeq \lim_{c \in Elt(X)} \text{Prsh}(C)/c \rightarrow \lim_{c \in Elt(X)} \mathcal{T}/F(c) \simeq \mathcal{T}/\widehat{F}(X) $$
(Note: one needs the fiber product preservation proved above to show this)
But as all the functors $\text{Prsh}(C)/c \rightarrow \mathcal{T}/F(c)$ preserves finite limits, their limits also preserve finite limits and hence for all $X \in \text{Prsh}(C)$ the functor $\text{Prsh}(C)/X \rightarrow \mathcal{T}/\widehat{F}(X)$ preserve all finite limits, which concludes the proof.