This is just a special case of Zsigmondy's Theorem, so the answer is $a=2$.

Starting with: $$ a^6 - 1 = (a^3 - 1)(a^3 + 1) = (a^3 - 1)(a+1)(a^2 - a + 1)$$Every prime divisor of $a^3 - 1$ is already prime divisor of $a^3 - 1$ (same numbers). So consider prime divisor of $a+ 1$ or $a^2 - a + 1$. Because every prime divisor of $a+1$ always divides $(a+1)(a-1) = a^2 - 1$, so only concern is $a^2 - a + 1$. Let $p$ be the prime divisor of $a^2 - a + 1$, and let $p|a^3 - 1$. Now, $p|a(a^2-a+1) - (a^3-1) = a^2 - a - 1$. Now $p|(a^2 - a +1) - (a^2 - a -1)$ which implicates $p=2$. But $2|(a(a-1)$ which means $2|1$, a contradiction. Now, let $p|a^2 - 1$, then $p|(a^2-a+1) - (a^2 - 1) = a - 2$. Here, $p|a(a-2) - (a^2-1) = 2a - 1$, which gives: $p|(2a-1) - 2(a-2) = 3$. Which means $p=3$. With this saying, only prime divisor of $a^2 - a + 1$ is $3$. Let, $$ a^2 - a + 1 = 3^\beta $$If $\beta \geq 2$ then $a^2-a+1 \equiv (mod9)$. This is impossible by checking all modules 9 of $a$. So, $a^2-a+1 = 3$ which gives $(a-2)(a+1)=0$, so only solution is $a=2$.