$p=2^n+1$. Prove that every quadratic nonresidue modulo $p$ is a primitive root modulo $p$

This is another one of the number theory problems I've been struggling with as of late (hopefully I'm not posting too many questions at once!).

Let $n$ be a positive integer and let $p=2^n+1$ be a prime number. Prove that every quadratic nonresidue modulo $p$ is a primitive root modulo $p$.

Suppose $a$ is a quadratic nonresidue and $r$ a primitive root muodulo $p$. Then there exists no $x$ such that $x^2 \equiv a \mod p$. We must determine the order of $a$. We know that $a$ must be congruent to some power of $r$, say $m$, since $\{r^1,r^2,\ldots,r^{2^n}\}$ is a reduced residue system modulo $p$. Suppose that $\mathrm{ord}_p a=b$. Then $a^b \equiv r^{mb} \equiv 1 \mod p$. We must show that $b=2^n$ since $\phi(p)=2n$. But $r$ is a primitive root and so $mb \equiv 0 \mod{2^n}$ (or $\equiv 2^n \mod{2^n}$). Hence $mb = 2^n k$. How can I finish it?

We will use standard results that you may be familiar with. There are $2^{n-1}$ quadratic non-residues of $p$.

There are $\varphi(\varphi(p))$ primitive roots of $p$. Thus there are $2^{n-1}$ primitive roots of $p$.

It follows that every quadratic non-residue is a primitive root.

If you know the basics of cyclic groups and their subgroups, then the following approach suggests itself.

The group $\Bbb{Z}_p^*$ is cyclic of order $p-1=2^n$. If $a$ is an element of order $2^n$, then it is primitive. OTOH if the order of $a$ is a factor of $2^{n-1}$, then it is a square, because the squares form the unique subgroup of order $2^{n-1}$. By the same argument a squre cannot be primitive.