Here is my reasoning:
$T^2$ has a cyclic vector iff its minimal polynomial and its characteristic polynomial are equal. Minimal polynomial of $T^2$ has degree $n$ . Therefore $T$ has an annihilating polynomial with degree $2n$ and with variables having even powers.
Since minimal polynomial of $T$ divides the annihilating polynomial having even powers, the powers of $x$ in minimal polynomial of $T$ will have the same parity. If minimal polynomial of $T$ is $q(x)$ and deg q(x) < n , $q(x)q(x)=p'(x^2)$ and p' is an annihilating polynomial of $T^2$ having deg < n which is a contradiction.
Is my reasoning is correct?