In Algebraic Extensions of Fields by Paul J. McCarthy (p. 68), the following definition is given:
A field $k$ is quasi-finite if it is perfect and if it has, in any algebraic closure, exactly one extension of degree $n$ for each integer $n > 0$.
So here it means that we don't even allow having two cubic extensions $F,F' \subset \overline k$ of $k$ such that $F \cong F'$ as $k$-algebras, but $F \neq F'$ ? (For instance $\Bbb Q(\zeta_3\sqrt[3]2) \cong \Bbb Q(\sqrt[3]2)$ as fields [hence as $\Bbb Q$-algebras, since the characteristic is $0$] even if these fields are not equal).
Or is it meant that we only require having only one extension of degree $n$, up to isomorphism? If so, it is up to "field isomorphism", or up to "$k$-algebra isomorphism"?
Even if we fix an algebraic closure of $k$, this is not clear to me what the precise condition should be.
If the definition is really "exactly one" (without even considering isomorphisms), then:
Is there a perfect field which is not quasi-finite but has exactly one extension of degree $n$, up to field isomorphisms (for any $n \geq 1$)?