So, here is my attempt:
Consider $d=(a,c)$. Per definition $d|c \Rightarrow d^n|c^n$. So by $c^n=ab$ we have $d^n|a$ since $(a,b)=1$. Therefore $a=k\cdot d^n$ for some $k$. I could argue the same way for $b=l\cdot e^n$ for some $l$ with $e=(c,b)$. So now we have $c^n=kld^n e^n$. Taking the nth root I have a whole number $c$ on the left, so the $k,l$ must b powers of n. Qed.
What do you say?
You have proven that $kl$ is an $n$-th power, not that $k$ and $l$ separately are $n$-th powers. So basically, you're back where you started ("$\gcd(k,l)=1$ and $kl=m^n$, what about $k$ and $l$?"), but with one important difference: unless $c=1$, we have $kl<ab$. This means you may tweak your proof into a proof by (strong) induction.
In the end, it will, of course, turn out that $k=l=1$. This suggests a second approach to completing your proof: Try showing that $c=de$.
Since $(a,b)=1$, we can write $a = p_1^{d_1} \cdots p_r^{d_r}$ and $b=q_1^{e_1} \cdots q_s^{d_s}$, where the $p_i$'s and $q_j$'s are distinct primes. By hypothesis $ab=c^n$, which implies that each exponent in $p_1^{d_1} \cdots p_r^{d_r} q_1^{e_1} \cdots q_s^{d_s}$ is a multiple of $n$. It follows that both $a$ and $b$ are $n$th powers.
If you have already established the fundamental theorem of arithmetic (unique factorization into primes) this is easy, since $a$ and $b$ cannot have a prime factor in common.
If you need a proof that doesn't use the FTA, @Arthur 's answer should help.
