IRL ABooklet

Our main tool will be the observation that
$$f(mn) = 1 \quad \text{implies} \quad f(m) = f(n) = 1. \qquad (6)$$
This is true because the functional equation $f(mn) + 1 = f(m) + f(n)$ then leads to $2 = f(m) + f(n)$ which is only possible when both positive integers, $f(m)$ and $f(n)$, are equal to 1. In particular, the given $f(2024) = 1$ and $2024 = 88 \cdot 23$ imply $f(23) = 1$. The given functional equation then implies
$$f(23n) + 1 = f(n) + f(23) = f(n) + 1,$$
hence $f(23n) = f(n)$ for all positive integers $n$. It follows by induction that $f(23^k n) = f(n)$ for all positive integers $k, n$.
If $n$ is not divisible by 23, the linear congruence $nx \equiv 22 \pmod{23}$ has a solution $x$ which can be chosen to be a positive integer. We then have $f(nx) = 1$ and so $f(n) = 1$ by (6).
We can write each positive integer as $m = 23^k n$ with some $k \ge 0$ and $n > 0$ not divisible by 23. From above we now have $f(m) = f(23^k n) = f(n) = 1$. Hence, the only function that satisfies the given condition is the constant function $f(m) = 1$ for all $m \in \mathbb{Z}_+$. It is easy to see that this function indeed satisfies all the conditions.