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

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.

A bit differently: With $r$ a primitive root, we can write $a\equiv r^m$ for some $m$. If $a$ is a quadratic nonresidue, clearly $m$ must be odd. Then $\gcd(m,2^n)=1$ and there exist $u,v\in\mathbb Z$ with $um+v2^n=1$. Thius implies $a^u=r^{um+v2^n}\equiv r$ and hence that $a$ is primitive.