 Solution

Let $g$ generate the cyclic group of nonzero residues, whose order $p-1=2^{2^n}$ is a power of two. Write $a=g^j$. Its multiplicative order is $(p-1)/\gcd (j,p-1)$, so it is a primitive root precisely when $j$ is odd. The quadratic residues are exactly the even powers of $g$. Thus the primitive roots are precisely the quadratic nonresidues, with nonzero residue classes understood.

For $n\ge 1$, $p\equiv 1(\mathrm{mod}4)$, so quadratic reciprocity gives $(7/p)=(p/7)$. Since $2^3\equiv 1(\mathrm{mod}7)$ and $2^n$ alternates between $1$ and $2$ modulo three, $p$ is respectively $3$ or $5$ modulo seven. The nonzero squares modulo seven are $1,2,4$. Consequently $(7/p)=-1$, andThere is a genuine endpoint exception if the printed form permits $n=0$: then $p=3$ and $7\equiv 1(\mathrm{mod}3)$, which has order one and is not primitive. The first equivalence remains true for $p=3$, but the final assertion requires excluding that prime.