Are given numbers $a,b$ fixed? I am confused about the statement of the problem. Because I have that $f(\lfloor{\sqrt{n}}\rfloor+b)$ divides $2\lfloor{\sqrt{n}}\rfloor+1$ numbers of the form $f(n+a)$ and if numbers $a,b$ are not fixed problem is trivial, otherwise I am a little stuck.

bump I am interested in a solution

Let's fix an arbitrary large positive integer $c > \max(b, a-b)$ for which $c+b^2-2b$ is divisible by all the primes which are $<2^{n+1}$. Then, observe that $f(k^2 + c)$ is divisible by $f(k+b)$ for big $k,$ since then we can just set $n = k^2 + c-a$. In view of this, let's define the sequence $\{a_i\}$ by $a_0 = k$, and $a_{i+1} = a_i^2 + c-b$ for $i \geq 0$, where $k$ is a big number to be chosen later. Then, we know that $f(a_i + b) | f(a_{i+1} + b)$ for all $i \geq 0$. Now, when are $a_0+b, a_1+b, \cdots, a_{n-1} + b$ pairwise relatively prime? Notice that for all primes $p$, $a_{i+1} \equiv a_i^2 + c - b$ (mod $p$). In view of this, let's define the sequence $\{x_i\}$ by $x_0 = 0$ and $x_{i+1} = x_i^2 + c-b$. Therefore, if we were to have that $p|a_i, a_{i+j}$ for some $j > 0, i \geq 0,$ then we would know that $p | x_j$. Let $p_1, p_2, \cdots, p_{t}$ be the primes which divide $x_1 \cdots x_{n-1}.$ Observe that if we were to find some $k = a_0$ such that $a_0 + b, a_1 + b, \cdots, a_{n-1} + b$ were all not divisible by any of the $p_i$'s, then we'd be clearly done, since no other prime $r$ could possibly divide two of them. We will show the existence of such a $k = a_0$ by CRT. Clearly, it suffices to show it for every $p_i$, by CRT. For $p_i < 2^{n+1}$, we know that $p_i|c-b,$ and so hence we may simply let $k \equiv a_0 \equiv 1-b$ (mod $p_i$), which would then ensure that $a_{i+1} \equiv a_i^2 + c -b \equiv (1-b)^2 + c - b \equiv 1-b$ (mod $p_i$) for all $i$ by induction, since $p_i | c+b^2 - 2b$ by how we chose For $p_i > 2^{n+1}$, observe that any nonconstant monic polynomial of degree at $2^{n+1} - 1$ must have some residue class mod $p_i$ which isn't a root of it. In particular, let's define the polynomials $P_0(x) = x$, and $P_{i+1}(x) = P_i(x)^2 + c - b$ for $i \geq 0.$ Then observe that $p_i \nmid a_r+b$ is equivalent to $p_i \nmid P_r(a_0) + b$. Therefore, if we consider some residue class $y$ which is not a root of $\prod_{i = 0} ^{n} (P_i(x) + b),$ we can just set $k = a_0 \equiv y,$ and be done. By CRT, and picking big enough $k$, we are done. $\square$