We can show that $f(n)\in\{1,2\}$ for all $n\geq 2$. Right? Futher, apply some reasonings to obtain that $f(pk+i)=1$ for $i=2,3,...,p$, $f(pk+1)=1$ or $f(pk+1)=2$ for all $k$, $f(1)$ is an arbitrary number. Or something like that. I am not sure. Excuse me all nonsenses I have written here.

Suppose $n \geq k$ and $p$ doesn't divide $n-1$. You see there is $k$ that $n-1\ |\ n+kp$. So $f(n)\ |\ f(n+kp)+1$ but we know that $f(n)=f(n+kp)$, Therefore $f(n)\ |\ 1$ and $f(n)=1$. Consider an arbitary $n \neq 1$, We know that $n-1\ |\ (n-1)kp$, So $f(n)\ |\ f\big((n-1)kp\big)+1=2$. So for every $n\neq 1,\ f(n) \in \{1,2\}$. Now we have 2 cases: a ) $f(n)=2, \forall n\geq k$ and $p\ |\ n-1$, Consider $n$ that $p$ doesn't divide $n-1$. There is $m$ that $n-1|m$ and $p|m-1$. So $f(n)\ |\ f(m)+1=3$ and $f(n)=1$. So $f(n)=1$ for all $n \geq k$ or $m$ doesn't divide $n$; And is arbitarily defined for every $n<k$ and $p\ |\ n-1$ b ) $f(n)=1, \forall n\geq k$ and $p\ |\ n-1$, In this case $f(n)=1, \forall n \geq k$, and if we suppose: $S=\{a\ |\ f(a)=2 \}$ then it doesn't exist $m,n \in S $ that $m-1\ |\ n$. So all of functions with the problem properties are defined like this: Suppose $S$ is a finite subset of natural numbers that there doesn't exist $m,n\in S$ that $m-1\ |\ n$ and for $n>1,\ f(n)=2$ iff $n\in S$. $f(1)$ is arbitarily defined with this condition that $f(2)\ |\ f(1)+1$

Note $f(px)=f(pk)$ for all $x\geq k$ now taking $m=pk-1,n=(pk)(pk-1)\implies f(pk)=1$ now take $m=px-1$ where $x<k$ and $n=pt(px-1)$ so again we get $f(px)=f(pk)=1$ for all $x\in\mathbb N$. Now taking $m=px,n=pkx$ we've $f(px+1)|2$ now so for $p|n-1$ we've $f(n)=1$ or $2$ now taking $(n-1,p)=1$ and by PHP there exist a large $k$ such that $n-1|kp-1$ now so taking $m=n-1$ we've $f(n)|f(n+kp)+1\implies f(n)=1$. So we've $f(n)=1$ for all $(n-1,p)\neq 1$ otherwise our $f$ is one or two. Now obviously we can't take $m,n$ such that $m\equiv 0 (mod p)$ and $n\equiv 1(mod p)$ and thus $f(px+1)$ can take any of one or two. Also $f(1)$ can be taken as arbitrary.

First let's deal only with large $m$: Case 1: $m \not \equiv 0 \pmod{p}$. Say $m \mid tp+1 \implies m \mid m+tp+1$ (it's obvious there exists such $t$). $P(m,m+tp+1): f(m+1) \mid f(m+tp+1)+1 \implies f(m+1) \mid f(m+1)+1 \implies f(m+1)=1$ (using $f(m+p)=f(m)$ for large $m$). Thus $f(m)=1$ for $m \not \equiv 1 \pmod{p}$. Taking $m=tp$ now (aka $m \equiv 0 \pmod{p}$), $P(tp,ltp): f(tp+1) \mid f(tlp)+1 \implies f(tp+1) \mid 2 \implies f(tp+1)=1$ or $2$. Thus $f(m)=1$ for $ m \not \equiv 1 \pmod{p}$, $f(m)=1, m \equiv 1 \pmod{p}$, and $f(m)=1$ are only 2 solutions for large $m$. Time to tackle small $m$: Case 1: Continuation of function $f(m)=1$ for $ m \not \equiv 1 \pmod{p}$, $f(m)=1, m \equiv 1 \pmod{p}$, $m$ large. Taking small $m \not \equiv 0 \pmod{p}$: $P(m,tm)(\text{gcd}(t,m)=1,t$ large):$f(m+1) \mid 2$. $P(m,tpm): f(m+1) \mid 3$. Thus $f(m)=1$ for small $m$, $m \not \equiv 1 \pmod{p}$. Taking $m=np \equiv 0 \pmod{p}$, $P(m,tm): f(np+1) \mid f(npt)+1 \implies f(m+1) \mid 2$. Thus $f(m)=1$ or $2$ (pointwise) for small $m \equiv 1 \pmod{p}$. Case 2: Continuation of function $f(m)=1$ for large $m$. Note $f(m+1) \mid f(tm)+1 \implies f(m+1) \mid 2 \implies f(m)=1$ or $2$ (pointwise). Thus our 2 solution sets, speaking broadly, are: 1. $f(n)=1$ for all $n \geq k$, $k$ chosen arbitrarily, $f(n)=1$ or $2$ for $n<k$, chosen arbitrarily for each $n$ (Clearly this works). 2. $f(n)=1$ for all $n \geq k$ where $n \not \equiv 1 \pmod{p}$, $f(n)=2$ for all $n \geq k$ where $n \equiv 1 \pmod{p}$, $f(n)=1$ or $2$ for all $n<k$, chosen arbitrarily. Again, this should work.

Possible source of error: I took this condition

to hold only for large $n$, as it was included in the same statement. It could be a misinterpretation which if fixed could make some of the pathological cases disappear, but I think it's intended as I interpreted it though...