the states $c^{\dagger}_1 c^{\dagger}_2 |0\rangle$ and $\psi^{\dagger}_1 \psi^{\dagger}_2 |0\rangle$ should be identical, up to a global phase. You can see that both of them are eigenstates of the Hamiltonian with the same energy $\epsilon_1 + \epsilon_2$.
You can see that if you act directly on the state with the Hamiltonian you'll get $$H c^{\dagger}_1 c^{\dagger}_2 |0\rangle = (\epsilon_1 + \epsilon_2)c^{\dagger}_1 c^{\dagger}_2 |0\rangle$$ (notice that the $\nu$-terms give zero as $c^{\dagger}_2 c_1 c^{\dagger}_1 c^{\dagger}_2$ has two $c^{\dagger}_2$ with no $c_2$ between them, so it will vanish, and similarly $c^{\dagger}_1 c_2 c^{\dagger}_1 c^{\dagger}_2$).
For the composition of the two single particle eigenstates you found, the energy of the doubly occupied state is the sum of the energies (as they are non-interacting) which again gives $\epsilon_1 + \epsilon_2$.
If you'll write the single states explicitly and carry the multiplication of them (taking care to normalize both of them) you'll see that $\psi^{\dagger}_1 \psi^{\dagger}_2 \propto c^{\dagger}_1 c^{\dagger}_2$
You can see that even without finding the expansion coefficients explicitly. Just by the notion of orthogonality we know that $$\psi^{\dagger}1 = e^{i\varphi}\cos(\theta) c^{\dagger}_1 + \sin(\theta) c^{\dagger}_2 \ \psi^{\dagger}_2 = \cos(\theta) c^{\dagger}_2 - e^{i\varphi}\sin(\theta) c^{\dagger}_1 $$ which is the only way to get $\left{ \psi_i, \psi^{\dagger}_j \right} = \delta{i,j}$, up to arbitrary phase assigned to the states (here I just chose the relative phase to always assigned to $c^{\dagger}_1$, but any other choice is valid). 66
So taking this form
$$ \psi^{\dagger}_1 \psi^{\dagger}_2 = \left[e^{i\varphi}\cos(\theta) c^{\dagger}_1 + \sin(\theta) c^{\dagger}_2\right]\left[\cos(\theta) c^{\dagger}_2 - e^{i\varphi}\sin(\theta) c^{\dagger}_1\right] = e^{i\varphi}\left[\cos^{2}(\theta) + \sin^{2}(\theta)\right]c^{\dagger}_1c^{\dagger}_2 = e^{i\varphi}c^{\dagger}_1 c^{\dagger}_2$$ which is what we wanted to show. The global phase of course doesn't matter at all. Note that here I used $c^{\dagger}_i c^{\dagger}_i = 0$ and also $c^{\dagger}_1 c^{\dagger}_2 = -c^{\dagger}_2 c^{\dagger}_1$. 66