Let $\alpha$ be such a root, then $\beta=\alpha^3$ satisfies $\beta^2+3=0$, hence ${1+\beta\over 2}={1+\alpha^3\over 2}$ is a $6^{th}$ root of $1$.
You can even verify this directly:
$$(1+\beta)^6 = 1+6\beta+15\beta^2+20\beta^3+15\beta^4+6\beta^5+\beta^6$$
Now use that $\beta^2=-3$ to reduce this to:
$$(1+5\cdot(-3)+15\cdot(-3)^2+(-3)^3)+\beta(6+20\cdot(-3)+6\cdot(-3)^2)$$
$$=(1-45+135-27)+\beta(6-60+54)= 64+0\beta $$
$$= 2^6.$$
And this is primitive as if we only cube things we get
$$(1+\beta)^3 = 1+3\beta + 3\beta^2+\beta^3 = 1-9 + \beta(3-3) = -8=-2^3.$$
First, recall Hensel's lemma:
Theorem: Let $a\in\Bbb Z_p$. If $f(X)\in\Bbb Z_p[X]$, $f(a) = 0\pmod{p}$ and $f'(a)\not\equiv 0\pmod{p}$, then there exists a unique $\alpha\in\Bbb Z_p$ such that $f(\alpha) = 0$ and $\alpha\equiv a\pmod{p}$.
This might not be quite the version you're used to, but it's essentially the same, except we're starting with a $p$-adic number and polynomial with $p$-adic coefficients. Existence and uniqueness will both be important for us in the following.
Remark: This version of Hensel's lemma answers both of your questions in the affirmative. If you have $\zeta = a_0 + a_1 p + \dots$ with $\zeta^m = 1$ in $\Bbb Z_p$ ($p\not\mid m$), and $a_0 = 1$, then $\zeta = 1$, since $1$ is already an element of $\Bbb Z_p$ satisfying $X^m - 1 = 0$, and thus uniqueness implies $\zeta = 1$. The answer to your second question is again yes by uniqueness, because $\zeta^t$ still satisfies $X^m - 1 = 0$ (and so does $1$) and $\zeta^t\equiv 1\pmod{p}$.
Here is the way I'd probably make the argument.
Lemma: There exists a primitive $p-1$st root of unity $\zeta\in\Bbb Z_p^\times$.
Note that this proves the existence portion of your question, because if $m>0$ divides $p - 1$, then an appropriate power of $\zeta$ will be a primitive $m$th root of unity.
Proof: Consider the polynomial $X^{p - 1} - 1\in\Bbb Z_p[X]$, and the elements $1,2,\dots, p - 1\in\Bbb Z_p$. Each of these elements satisfies $a^{p - 1} - 1\equiv 0\pmod{p}$. Moreover, $-a^{p - 2}\not\equiv 0\pmod{p}$, so for each of the elements we considered we obtain an $\alpha\equiv a\pmod{p}$ with $\alpha^{p - 1} - 1 = 0$. So we have found $p - 1$ $p - 1$st roots of unity. To see that one is primitive, recall that a finite subgroup of the multiplicative group of a field is cyclic, and the elements in $\Bbb Q_p^\times$ satisfying $X^{p - 1} - 1 = 0$ form a subgroup. Those elements are precisely the $p - 1$ $\alpha$'s we found, because a degree $n$ polynomial over a field can have at most $n$ roots.
Now, we must show that if $m$ does not divide $p - 1$, there is not a primitive $m$th root of unity.
Lemma: Let $m > 0$ be a positive integer such that $p\not\mid m$. Suppose that $\alpha\in\Bbb Z_p^\times$ satisfies $\alpha^m = 1$. Then $\alpha$ satisfies $\alpha^n = 1$ for some $n\mid p - 1$.
Proof: (adapted from the notes in my above comment) Suppose $\zeta\in\Bbb Z_p^\times$ satisfies $\zeta^m = 1$. Then $\zeta\not\equiv 0\pmod{p}$, so there exists some $p - 1$st root of unity $\alpha$ with $\alpha\equiv\zeta\pmod{p}$ by our argument for the last lemma. Then $\alpha$ and $\zeta$ both satisfy $f(X) = X^{m(p - 1)} - 1 = 0$. (Note that $f'(\alpha),f'(\zeta)\not\equiv 0\pmod{p}$.) But now we have two elements of $\Bbb Z_p$ satisfying the same polynomial $f\in\Bbb Z_p[X]$ which are equivalent modulo $p$, so by uniqueness in Hensel's lemma, $\zeta = \alpha$. Thus, $\zeta$ is a $p-1$st root of unity, so it is a primitive $n$th root only for some $n$ dividing $p - 1$.
Note that we did not need to assume $\alpha$ is a primitive $m$th root for the last argument. The lemma holds for any $m$th root, and will thus hold for a primitive $m$th root.
Best Answer
In a word, $\Bbb Q_p$ has no $p$-th roots of unity, except for $p=2$. Beyond that, the torsion in the multiplicative group is exactly isomorphic to the multiplicative structure of $\Bbb F_p$: cyclic of order $p-1$. So, to answer your question, you only need look at the relationship between $m$ and $p-1$.