Why do people say that neutrinos are either Dirac or Majorana fermions?

You're completely correct: it's perfectly allowed to have both Dirac and Majorana mass terms. However, the presence of a Majorana mass term (whether or not a Dirac mass term is present) implies the violation of lepton number. When people say they're testing for whether a neutrino is Majorana, they just mean that they're looking for such violations. For a nice review of some simple neutrino mass models, phrased in the same terms that you used, see the relevant chapter in Burgess and Moore, The Standard Model.

I don't think this necessarily is sloppy language. I think that in condensed matter, whether a fermion is Majorana or not is a sharply defined, important thing. However, in particle physics, when we say that a particle is a Blah fermion (where Blah could be Weyl, Majorana, or Dirac), we mean that we have in mind a description for that particle in terms of Blah fermion fields.

For example, a given massless neutrino state could be created by a left-chiral Weyl field, a right-chiral Weyl field, or a Majorana field. None of this affects the physics; the fields are just a bookkeeping tool that help us write down interactions for the particles. As a more extreme example, Burgess and Moore go further and describe all the fermions in the Standard Model as Majorana fields (i.e. the electron corresponds to two separate Majorana fields, but with their Majorana mass terms each set to zero), solely because this allows them to use 4-component spinors and the associated computational tools.

Historically, the distinction between Weyl, Dirac, and Majorana fields was based on the fields' Lorentz transformation properties. However, these days this is becoming less important, so the same words are repurposed. In condensed matter, the words' original meanings can't matter because there's no Lorentz symmetry, so they seem to be used to denote properties of the spectrum, or of the (anti-)commutation relations describing the system. And in particle physics, the original meanings are less important in neutrino physics for the reasons I gave above, so they are adapted to pin down the only physical thing that varies between the possibilities -- namely, whether the particle number is conserved.

I have to say that I do not completely agree with the answer given by knzhou as I think he misses a crucial point in his explanation.

Of course it is right that the most general mass term contains both Dirac and Majorana terms and the appearance of Majorana terms implies lepton number violation. We can summarize the mass term in matrix form as $$-\mathcal{L}_m = \frac{1}{2}n_L^TC\mathcal{M}n_L + h.c.$$ with $$n_L = \left(\begin{matrix}\nu_L\\(N_R)^c\end{matrix}\right)$$ and $$\mathcal{M}=\left(\begin{matrix}M_L & M_D \\ M_D^{T} & M_R\end{matrix}\right)\label{eq:neutrino_mass_matrix}$$ Here, $M_D,M_L$ and $M_R$ are $n\times n$ matrices (where n is the number of generations) and represent Dirac mass terms, left-handed Majorana mass terms and right-handed Majorana mass terms.

So far so good. But we should not miss one point. Here we are looking at neutrinos as flavor states. When talking about massive particles we have to diagonalize the mass matrix. Assuming $M_R$ to be invertible, we can block-diagonalize by a base transformation $$-\mathcal{L}_m\longrightarrow\frac{1}{2}\chi_L^TC\mathcal{M}_{\rm{diag}}\chi_L + h.c.$$ with $$ n_L=U\chi_L\\\mathcal{M_{\rm{diag}}}= U^T\mathcal{M}U = \left(\begin{matrix}\tilde{M}_L & 0 \\ 0 & \tilde{M}_R\end{matrix}\right)$$ now we are left with massive fields $\chi_L$ which have only a Majorana mass term.

You can do the whole calculation in the limit of 1 generation to check.

This is nicely explained in the lectures on neutrino physics by Evgeny Akhmedov.