I'm doing and writing this while I'm a bit drunk/sleepy, so my apology if it's wrong. Please let me know if it is so. It's enough to find such $(a,b)$ that satisfies the condition when $n$ is a prime power, say $n=p^k$ for prime $p$ and positive integer $k$. Choosing $k=1$ gives $p^a+1\mid p^b+1\implies b=ta$ for some odd number $t$. Now, we have $f_a(p^k)=\frac{p^{(k+1)a}-1}{p^a-1}$ and $f_b(p^k)=\frac{p^{(k+1)ta}-1}{p^{ta}-1}$. For any odd prime $r$ that $r\mid f_a(p^k)$, let $\ell =\mathrm{ord}_r (p)$. Note that $\ell \mid (k+1)a$. We get that $$\nu_r(f_a(p^k)) =\begin{cases} (\nu_r(p^{\ell} -1)+\nu_r(\frac{(k+1)a}{\ell}))-(\nu_r(p^{\ell} -1)+\nu_r(\frac{a}{\ell}))=\nu_r(k+1), & \text{ if }r\mid p^a-1 \\ \nu_r(p^{\ell} -1)+\nu_r(\frac{(k+1)a}{\ell}), & \text{ if }r\nmid p^a-1 \end{cases}.$$The similar relation holds when $a$ is replaced by $b$. If $t\geqslant 3$, i.e., $b>a$. Consider $p=5$ and let $r$ be a prime number that divides $p^b-1$ but doesn't divide $p^t-1$ for any $1\leqslant t<b$. Also, choose $k+1=b$. We get that $$r\nmid p^a-1\implies \nu_r (f_a(p^k)) =\nu_r(p^{\ell} -1)+\nu_r\left( \frac{(k+1)a}{\ell}\right)=\nu_r (p^b-1)+\nu_r (a),$$while $r\mid p^{ta}-1\implies \nu_r(f_b(p^k))=\nu_r (k+1)=\nu_r (b)$. But since $r\mid p^{r-1}-1$, we get that $b\leqslant r-1\implies \nu_r(a)=\nu_r (b)=0$. So, $\nu_r (f_a(p^k)) <\nu_r (f_b(p^k))$. Contradiction. Hence, the only answer is $(a,b)=(c,c)$ for every positive integer $c$.

ThE-dArK-lOrD wrote: I'm doing and writing this while I'm a bit drunk... That's dangerous; never drink and derive.

If $t>1$ you can set $k=t-1$ and get a contradiction through LTE.

Lamp909 wrote: If $t>1$ you can set $k=t-1$ and get a contradiction through LTE. Elaborate please.

ThE-dArK-lOrD wrote: I'm doing and writing this while I'm a bit drunk/sleepy, so my apologize if it's wrong. Please let me know if it's wrong. Total bro,someone wants to fight when they're drunk,someone wants to sleep,you wanna solve problems,now that's dedication!

Wait is below wrong if not it is quite short solution . It is clear that $b=ak $ where k is odd integer. If $k\neq 1$ then take $n=p^{q-1} $ where $q|k $ .This easily gives contradiction. So $b=a $ Note:I apologize if this is wrong or same with ThE-dArK-lOrD's (I didnot read his solution)

tenplusten wrote: Wait is below wrong if not it is quite short solution . It is clear that $b=ak $ where k is odd integer. If $k\neq 1$ then take $n=p^{q-1} $ where $q|k $ .This easily gives contradiction. So $b=a $ Note:I apologize if this is wrong or same with ThE-dArK-lOrD's (I didnot read his solution) Your solution is absolutely correct and concise,my friend(coincides with my solution).It seems people can find easy solutions when they are not drowsy!

tenplusten wrote: If $k\neq 1$ then take $n=p^{q-1} $ where $q|k $ .This easily gives contradiction. So $b=a $ Can someone please explain why we get a contradiction?

aq th roots are contained in ak th roots