If $a^{n}-1$ is prime then $a=2$ and $n$ is prime?

The proof is alright there are two or three details though (the same issue twice actually), oe was already pointed out in comments:

• Likely you should exclude the case $a=1$ right away. Just by saying $1^n -1 = 0$ is not prime so assume $a>1$.

• You cannot derive from $a^n -1$ being a prime and $(a-1) \mid (a^n-1)$ directly that $(a-1)= 1$. What you can do is say $a-1=1$ or $a-1 = a^n-1$. The latter is impossible as $n > 1$ (and $a >1$); note here you use $n>1$.

• When you assert that $a^{kl}-1$ is composite, you should state explicitly that both factors you exhibit are not $1$.

These are not major problems, but if one is picky one could insist on them.