As Mark says, nothing is known about infinitely many primes in dynamical sequences, other than ones that contain an arithmetic progression. An easier, but still useful, question, is that of primitive prime divisors. A prime $p$ is a *primitive prime divisor* of $f^n(x)$ if $p\mid f^n(x)$ and $p\nmid f^m(x)$ for all $m\lt n$. There's a vast literature on primitive prime divisors in various sorts of sequences. In general, if $\mathcal{A}=(a_n)$ is a sequence of integers, or more generally, rational numbers, the *Zsigmondy set* of $\mathcal{A}$ is
$$
\mathcal Z(\mathcal A) = \{ n : \hbox{the numerator of $a_n$ has no primitive prime divisors} \}.
$$
Patrick Ingram and I [1] showed that under suitable hypotheses on $f\in\mathbb{Q}(T)$ and $x,y\in\mathbb{Q}$, the Zsigmondy set $\mathcal{Z}(f^n(x)-y)$ is finite. In addition to some obvious conditions needed to avoid trivial counterexamples, we needed to assume that $y$ is preperiodic for $f$. Recently, Gratton, Nguyen, and Tucker [2] removed this restriction on $y$, conditional on the $abc$ conjecture.

[1] Ingram, Patrick; Silverman, Joseph H.; Primitive divisors in arithmetic dynamics. *Math. Proc. Cambridge Philos. Soc.* **146** (2009), no. 2, 289–302 (MR2475968)

[2] Chad Gratton, Khoa Nguyen, Thomas J. Tucker, ABC implies primitive prime divisors in arithmetic dynamic, preprint, http://arxiv.org/abs/1208.2989

Pedantic comment alert: Presumably this is not exactly what you meant: $f(x)=2x-7$ satisfies $f^n(7)$ is prime for all $n$. $\endgroup$`$\{n : f^n(x)\in Y\}$`

is a finite union of arithmetic progressions. Here $f:X\to X$ is a morphism of a (smooth) projective variety, and $Y\subset X$ is a (smooth) subvariety. $\endgroup$