If $c=1$, we can take the sequence of all primes, which is infinite. If $c>1$ and a positive integer $n$ satisfies $f(n)=n+c$, then $n$ is not a prime. Hence it has at least one divisor $d$ with $\sqrt{n} \leq d < n$. In particular, $f(n) > \sqrt{n}+n$ and hence $\sqrt{n} < c$. Thus there can be at most finitely many such numbers. Consequently the answer is $\boxed{c=1}$.