We cannot apply the result of decomposition of $x^4-1$ and $x^6-1$ because $\gcd(2,4\ \text{or} \ 6)=2\neq 1$. That means that there are only $2$ binary cyclic codes for each $n=4$ and $n=6$ ?
Edit : in $\mathbb{F}_2$, we have $x^6-1= (x+1)^2(x^2+x+1)^2$ so for $n=6$ we will have $9$ binary cyclic codes.
And for $n=4$, $x^4-1=(x+1)^4$ we will have $5$ binary cyclic codes.
Thanks in advance !