Consider $P^f_n=|\{m<n|f(m)\in\mathbb P\}|$, where $f$ is a strictly increasing function $\mathbb N^+\to \mathbb N^+$. Below a diagram of $P^f_n$ for different functions $f$:
(Are there polynomials $q$ which are strictly increasing on $\mathbb N^+$ such that $P^q_n>>P^u_n$, where $u(n)=2n-1$?)
Changed question:
Is there an upper bound strictly less than $p_n$?
Is there an optimal polynomial of degree 1?
I first believed that the polynomial $n\curvearrowright k!\cdot n+1$ would give larger $P^f_n$ for increasing $k$, which isn't true. The most high density polynomial I found so far is $n\curvearrowright 30n-7$
Suppose there is a linear function $f$ such that for all linear functions $g$ and all $n,m>N$ for some integer $N$ we have $$\frac{P^f_n}{n}\ge\frac{P^g_m}{m}$$
Write $f(x)=ax+b$. Now, for all positive integers $n>N$ ($m$ is also assumed to be integer) $$\{0<m<bn:am+b\text{ is prime}\}=\bigcup_{i=0}^{b-1}\{0<m<n:abm+ai+b\text{ is prime}\}$$ On the right side there are $b$ sets. However, the one with $i=0$ does not contain any primes, which means there is at least one $0\le k\le b-1$ such that $$\{0<m<n:abm+ak+b\}$$ contains more than $\frac 1b P^f_{bn}$ primes. Defining $g(x)=abx+ak+b$ gives: $$\frac1bP^f_{bn}<P^g_n$$ dividing both sides by $n$ yields: $$\frac{P^f_{bn}}{bn}<\frac{P^g_n}{n}$$ which contradicts our earlier statement about $f$, meaning that there does not exist a linear function $f$ such that for all linear functions $g$ and natural numbers $n,m$ larger than some constant $N$ we have $$\frac{P^f_n}{n}\ge\frac{P^g_m}{m}$$