(I meant to write: the product over the closed points; that is to say, those of finite residue field $k(x)$).

By the way, the Hasse-Weil zeta function of a finite type scheme $X_{\mathbb{Z}}$ over $\mathbb{Z}$ (the integral model, that is), is simply the product $\prod_x (1 - |k(x)|^{-s})^{-1}$ over the closed space. But since there is no $\mathbb{Z}$-model in general, what you do instead is to take your $P_{i,p}$ to be the char. poly. of $\mathrm{Frob}_p$ acting on $H^i(\overline{X},\mathbb{Q}_l)^{I_p}$, the invariants of inertia. This does not refer to a $\mathbb{Z}$-model, and works to define the L-function of an abstract Galois representation. See this: ucl.ac.uk/~ucahmki/ihes3.pdf

This theorem of Soler is just great! Thank you for this link!

Alternatively, take a look at the proof of Thm. 12.2.15 in Bombieri-Gubler. The conjecture is shown to be equivalent to the statement that the set of $n$ for which there is a prime $p \geq n$ with $p^2 \mid f(n)$ has density 0. For $\deg{f} \leq 2$, this is obvious. When $f$ splits into such polynomials, you get a finite union of sets of density 0.

Take a look at the posting of Ravi B on this link: artofproblemsolving.com/Forum/… . For k-frees, the argument works assuming $f$ splits into factors of degree at most k. As for the Browning result you quote, my guess is that the proof would likewise work assuming $f$ splits into irreducibles whose degrees satisfy the mentioned inequality.

I mean, $E = \phi^* H$.

Saying that $E$ is linearly equivalent to $D$ means that you can find a hyperplane $H$ in $\mathbb{P}^N$ such that $E = f^*H$. Now, either $H$ contains the point $\phi(C)$, or it doesn't.