Plugging in $n=1$ gives $f(2)f(3)=12$, therefore $(f(2), f(3))=(2,6)$ or $(3,4)$. However, the former implies $$f(4)f(5) \geq (6+1)(6+2)>42=9 \cdot 2^{2}+3 \cdot 2$$ which is impossible; therefore $f(2)=3$ and $f(3)=4$. We now show by induction with step size 2 that $f(2n)=3f(n)$ and $f(2n+1)=3f(n)+1$ for all $n$; the base case $n=1$ has already been proven. Assume the statement is true for $n<2k$. Applying the given and the inductive hypothesis, we have $$\begin{aligned} f(4k)f(4k+1) & =(3f(2k))(3f(2k)+1)=(9f(k))(9f(k)+1) \\ f(4k+2)f(4k+3) & =(3f(2k+1))(3f(2k+1)+1)=(9f(k)+3)(9f(k)+4) \end{aligned}$$ Let $x=f(4k+1)$. Since $f$ is strictly increasing, this implies $x \geq \sqrt{f(4k)f(4k+1)}>9f(k)$ and $x \leq \sqrt{f(4k+2)f(4k+3)}-1<9f(k)+3$. So $x=9f(k)+1$ or $x=9f(k)+2$. Since $9f(k)+2$ does not divide $9f(k)(9f(k)+1)$, we must have $f(4k+1)=x=9f(k)+1$ and $f(4k)=9f(k)$. A similar argument shows that $f(4k+2)=9f(k)+3$ and $f(4k+3)=9f(k)+4$, and this completes the inductive step. Now it is a straightforward induction to show that $f$ is the function that takes a number's binary digits and treats it as base 3. Since $137=10001001_{2}$ in binary, $f(137)=10001001_{3}=3^{7}+3^{3}+1=2215$.