If $l$ is a prime that divides $F_n$, then $2^{2^n}\equiv -1 \pmod{l}$, and therefore $2^{2^{n+1}}\equiv 1 \pmod{l}$. So $2$ has order $2^{n+1}$ modulo $l$. Equivalently, but in more group-theoretic language, (the equivalence class of) $2$ has order $2^{n+1}$ in $(\mathbb{Z}/l\mathbb{Z})^\times$. (The order of $2$ must divide $2^{n+1}$, but cannot be $\le 2^n$.)
If $p$ is a prime that divides $2^{2^k}+1$, then by the same reasoning $2$ has order $2^{k+1}$ modulo $p$. If $k<n$, that means that $p$ cannot be equal to $l$. So we have proved that no prime that divides $F_n$ can divide any $F_k$ with $k<n$. This shows that $F_n$ and $F_k$ are relatively prime.
Remark: We do not really need to identify the order of $2$ explicitly. For if $l$ is a prime that divides $F_n$, then $2^{2^n}\equiv -1 \pmod{l}$. But if $p$ is a prime that divides $F_k$ for some $k<n$, then $2^{2^k}\equiv -1\pmod{p}$, and therefore $2^{2^{k+1}}\equiv 1\pmod{p}$. Since $k+1 \le n$, this is impossible.
How to argue depends on what you know about coprime integers.
I.
For example, if you know that $a,b$ being coprime means that there are integers $n,m$ with
$$ an+bm=1, $$
you can argue as follows:
First, if $a^2$ and $b^2$ are coprime, as witnessed by integers $n,m$, so
$$ a^2n+b^2m=1, $$
then $k=an$ and $l=bm$ witness that $a,b$ are coprime:
$$ ak+bl=1. $$
The other direction is more interesting: First, note that if $x,y$ are coprime, so are $x^2,y$:
$$ xn+ym=1\Longrightarrow (xn+ym)^2=1, $$
but $$ (xn+ym)^2= x^2 n^2+y(2xnm+ym^2)= x^2k+yl, $$
with $k=n^2$ and $l=2xnm+ym^2$.
This gives us that if $a,b$ are coprime, $a^2,b$ are coprime. But then, letting $x=b,y=a^2$, this gives us that $x^2=b^2$ and $y=a^2$ are coprime as well, as we wanted.
II.
On the other hand, if you know that being coprime can be verified in terms of common prime divisors, then the argument is simpler, because if $p$ is prime and $p|xy$ then $p|x$ or $p|y$. This means that for any prime $p$, we have that $p|a^2$ iff $p|a$, and the same with $b^2$ and $b$, so the prime divisors of $a^2,b^2$ are the same as those of $a,b$. Hence, one pair is coprime iff the other is coprime.
Best Answer
For contradiction, assume $\gcd(a,b)=1$, but $\gcd\left(a^k,b^l\right)>1$. Let $p\mid \gcd\left(a^k,b^l\right)$ for some prime $p$. Then $p\mid a^k, b^l$. Also $p\mid a^k\implies p\mid a$ (by Euclid's Lemma) and similarly $p\mid b^l\implies p\mid b$. Then $p\mid a,b\implies \gcd(a,b)\ge p>1$, contradiction.