Here is an indirect solution - a solution that doesn't use any specific primitive root. I think a better solution exists, but I would still like to post it because it gives a different perspective.
The solutions $1$ and $m-1$ are obviously always solutions. So the difficulty is proving there aren't any more.
By the primitive root theorem, the possibilities for $m$ are $m=2$, $m=4$, $m=p^k$, or $m=2p^k$ where $p$ is an odd prime and $k\ge 1$.
For $m=2$ we actually have $1=m-1$, and there is a unique square root of $1$.
For $m=4$ you may check directly that the claim holds, just by calculating.
For $m=p^k$, we prove by induction on $k$. For $k=1$ there are no zero divisors, so $(x-1)(x+1)=0$ implies $x-1=0$ or $x+1=0$. For $k>1$, assuming there are exactly two solutions modulo $p^{k-1}$, we use Hensel's lemma to claim that these lift to exactly two solutions modulo $p^k$.
For $m=2p^k$ we use the Chinese remainder theorem to combine the two solutions modulo $p^k$ with the unique solution modulo $2$ to get two solutions modulo $m$.
Let $g$ be an primitive root mod $m$, then the set $S=\{g,g^2,\cdots, g^{\phi(m)}\}$ forms a reduced residue set mod $m$.
Since $(a,m)=1$, we can express $a$ as a power of $g$, let $a\equiv g^k\pmod m$. So by assumption we have $$g^{k\phi(m)/2}\equiv 1\pmod m.$$
But $g$ is primitive, we see $\phi(m)|k\phi(m)/2$ which is $2|k$, this shows that $a$ is an even power of $g$, which is equivalent to say that $a$ is a quadratic residue modulo $m$.
Best Answer
Let $m\geq 3$ and $g$ is a primitive root modulo $m$. Then $\mathbb Z^*_m$ can be fully represented by $\{g_0,\dots,g^{\varphi(m)-1}\}$. Hence, any $x\in \mathbb Z^*_m$ can be written as $x=g^k$ for some $0\leq k<\varphi (m)$ and working in $\mathbb Z_m$: $$x^2=1 \Leftrightarrow (g^k)^2= 1 \Leftrightarrow g^{2k}= 1 \Leftrightarrow \varphi(m) \mid 2k$$ which is only possible for at most two values of $k$, namely $0$ and $\frac{\varphi(m)}{2}$. As $\pm 1$ are always solutions and distinct, those are exactly the two solutions.
The other direction is most likely more difficult, I don't have a proof yet, but you can of course deduce this from Euler's theorem:
If there is no primitive root modulo $m$, then by Euler's theorem $m$ can not be of the form $$2, 4, p^k, 2p^k$$ where $p$ is an odd prime and $k$ is a positive integer. Check carefully, that this implies that $m$ is either a power of two or can be written as product of two coprime numbers which are both at least $3$. In both cases we find more than two solutions: