Suppose that we are really going to do it by induction, perhaps because a homework exercise specifies that we must. There are quite a few better ways than induction for solving this problem, most of which you will learn from the other answers. But let's do induction.
We experiment a bit. Start say at $n=1$, or even that mathematicians' favourite, $n=0$.
If $n=1$, then $2^{3n}=7$. Let $n=2$. Then $2^{3n}=63$. Note that $63$ is divisible by the primes $3$ and $7$. Let $n=3$. Then $2^{3n}-1=511$. But $511=7\times 73$. All our numbers so far are divisible by $7$. Let $n=4$. Then $2^{3n}-1=4095$. Is it divisible by $7$? The calculator says yes.
So maybe all of our numbers are divisible by $7$. Since all the numbers for $n\ge 2$ are $>7$, this will show that if $n\ge 2$, then $2^{3n}-1$ is composite. So we try to prove, by induction on $n$, that $7$ divides $2^{3n}-1$ for every positive integer $n$.
Certainly it is true at $n=1$. Suppose that we know that for a particular $k$, $2^{3k}-1$ is divisible by $7$. Can we prove that $2^{3(k+1)}-1$ is divisible by $7$?
Note that $2^{3(k+1)}-1=(2^{3k}-1)+(2^{3(k+1)} -2^{3k}).$
By the induction hypothesis, $2^{3k}-1$ is divisible by $7$. If we can prove that $2^{3(k+1)} -2^{3k}$ is divisible by $7$, we will be finished.
Note that $2^{3(k+1)} -2^{3k}=2^{3k+3}-2^{3k}=2^{3k}(2^3-1)=7(2^{3k}),$ so we are finished.