A Increasing Multiplicative Functional Equation where $nm$ is a cube 
Let $f:\mathbb{N}\rightarrow\mathbb{N}$ be a strictly increasing
function such that $$f(2)=2$$ and $$f(mn)=f(m)f(n)$$ for all positive
integers $m,n$  such that $mn$ is a perfect cube.
Prove that $f(n)=n$ for every positive integer $n$.

This problem was inspired by the $24$th W.L. Putnam Mathematical Competition, whose proof can be seen here on page 8.
Here is my attempted proof.
Note that $f(9)f(192)=f(24)f(72),\quad f(3)f(72)=f(24)f(9)$ .
Multiplying these two equations give that $f(24)^{2}=f(3)f(192)$.
Note that $f(24) \le f(9)f(3)-3$, and $f(192) \ge f(3)f(9)+165$.
This gives us that $f(3)^2(f(9)^2-f(9))-f(3)(6f(9)+165)+9 \ge 0$.
However, this route proved unsuccesful. I had tried to show that $f(24)=24$, which would imply by induction that $f(n)=n$ for all $n$.
Is there a way to easily solve the problem above? Any help would be appreciated.
 A: Because it is strictly increasing, all we have to do is prove there are an infinite number of solutions to $f(n) = n$, since all the numbers in between two solutions are also solutions themselves.  Using induction on $f(8 \cdot x^3) = f(8) \cdot f(x^3)$, the problem reduces to showing there exists $n \ge 4$ with $f(n)=n$.  It looks like you got at least this far with your $f(24)$ idea.  Below is my attempt:
We have $f(8^k) = f(8)^k$ and also $f(27^k) = f(27)^k$ for all $k$.  So for any $x$ which is a power of $8$, $a = \frac{\log{f(8)}}{\log{8}}$, $f(x) = x^{a}$ and for any $x$ which is a power of $27$, $b = \frac{\log{f(27)}}{\log{27}}$, $f(x) = x^{b}$.  For all $t$ there is an $s$ such that $8^s \lt 27^t \lt 8^{s+1}$, and since $f$ is increasing, $f(8^s) \lt f(27^t) \lt f(8^{s+1})$, which expands to $(8^{s})^{a} \lt (27^{t})^{b} \lt (8^{s+1})^{a}$.  But this means $a = b$.  For example, if $b \lt a$, then for sufficiently large $t$ it will be the case that $(27^t)^b \lt (8^s)^a$, which contradicts $f$ being increasing.  This is because $27^t$ is no more than $8$ times larger than $8^s$.  Likewise, assuming $a \lt b$ leads to a contradiction.  All of the above works for any two cubes instead of $8$ and $27$, so by transitivity, there exists  $c \ge 3$ such that for all $x$, $f(x^3) = x^c$.
Now, $f(125) = 5^c$ and $f(128) = \frac{f(8)}{f(4)} \cdot f(64) = 2 \cdot 4^c$.  Using $f(128) \gt f(125)$, $3 \le c \lt \frac{\log{2}}{\log{5} - \log{4}} \approx 3.10628 \lt 4$.  So $c = 3$, since it must be an integer for $f(x^3) = x^c$ to always be an integer, and that proves the proposition. 
