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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?

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    A simpler argument is $V=Z(v,T^2) \subseteq Z(v,T)$ implies $V=Z(v,T)$ . – 2017-02-21
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    @lhf : I saw this answer but I wanted to know if its possible to use the minimal polynomial argument. – 2017-02-21
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    *"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."* Is this a well known statement about polynomials? – 2017-02-21
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    How can you guarantee that $q(x)q(x)$ will be of the form $p'(x^2)$? – 2017-02-21
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    Let's suppose for now that I accept your explanation for the first comment. I'm still skeptical about that second one. $q(x) = x^3 + 1$ leads to $q(x)q(x) = x^6 + 2x^3 + 1$, which is not a function of $x^2$. – 2017-02-21
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    @ Omnomnomnom Anyway I saw my mistake. (x^2 -x)(x^2 +x) has odd powers cancelling out which I didn't take into consideration. isn't 1 x^0 which is even? Thanks for your help – 2017-02-21
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    Yes, $1$ is a degree zero polynomial – 2017-02-21
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    @Omnomnomnom I just got an idea based on my counterexample. Suppose $q(x)$ is the minimal polynomial of $T$ and that deg q < n. Then I define $q'(x)$ as a polynomial equal to $q(x)$ except that the sign of coefficients of odd powers of $x$ is opposite. I think that the product $q'(x)q(x)$ will have all powers even and its degree is less than $2n$. Therefore if $p'(x^2) = q'(x)q(x)$, $p'(x)$ will be an annihilating polynomial of $T^2$ with degree less than n. So $T^2$ can't be cyclic. – 2017-02-22
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    In other words, you've taken $p'(x)=p(-x)$. I think that works. I don't understand why you insist on trying to do things this way, though. – 2017-02-22
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    What about the converse statement? I am having trouble figuring this out :/ – 2018-10-15

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I don't understand why go through all the trouble. Vector $x$ is cyclic for $T$ if vectors $T^nx$ span whole space. Clearly, vectors of the form $T^{2n}x$ are a subset of these.