Let $p \equiv 1 \bmod 8$ be a prime number, let $K = {\mathbb Q}(\sqrt[4]{p})$, and let $F$ be the quartic subfield of the field of $p$-th roots of unity. An easy exercise involving Abhyankar's Lemma shows that $FK/K$ is an unramified quadratic extension, hence the class number of $K$ is always even.

The field $KF$ has the quartic subfield $L = {\mathbb Q}(\sqrt{u})$, where $u$ is the fundamental unit of $k = {\mathbb Q}(\sqrt{p})$. An routine application of the ambiguous class number formula to $L/k$ shows that $L$ has odd class number (there are two ramified primes, one infinite and the other one above $2$; clearly $-1$ is not a norm residue at the infinite prime).

Now I claim that if $p \equiv 9 \bmod 16$, the class number of $KF$ is odd. By class field theory, this implies that the $2$-class number of $K$ must be $2$. An application of the ambiguous class number formula to $KF/L$ shows that the $2$-part of the ambiguous class group has order
$$ h = \frac{2}{(E:H)}, $$
where $E$ is the unit group of $L$ and $H$ its subgroup of units that are norms from all completions of $KF$: in fact, only the two prime ideals above $p$ are ramified in $KF/L$. Thus it is sufficient to show that $E \ne H$. I will show that $\sqrt{u}$ is a quadratic nonresidue modulo the primes $\mathfrak p$ above $p$. But if $u = T + U \sqrt{p}$ (replace $u$ by $u^3$ in order to guarantee that $T$ and $U$ are integers), then $(\sqrt{u}/{\mathfrak p})_2 = (u/\mathfrak p)_4 = (T/p)_4 = (T^2/p)_8 = (-1/p)_8 = -1$
because $p \equiv 9 \bmod 16$; here we have used the congruence $T^2 \equiv -1 \bmod p$.

The reason why the case $(2/p)_4 = -1$ is easier is because in this case, the ideal above
$2$ ramified in $K$ generates a class of order $2$ in the $2$-class group, whereas this prime generates a class with odd order if $(2/p)_4 = +1$, which means that there is no strongly ambiguous ideal class in this case.

**Edit.** Paul Monsky has kindly written up this argument, filled in all the details, and made it available here. Thanks!