Obtaining wave function from field equation

A wave function is the projection of a state in the coordinate basis, where the coordinate basis is defined as the basis that diagonalizes the position operator ${hat x}$. In QFT, the analog of this is the fundamental field operator ${hat Psi}$ itself. Eigenvalues of this are defined as $$ {hat Psi} (t,vec{x}) | phi rangle = phi(vec{x}) | phi rangle $$ Then, given a state $| alpha rangle$, its wavefunction is defined $$ alpha [ phi (vec{x})] = langle phi | alpha rangle~. $$ Remember that in QFT, the wave function is really a wave-functional.

Weinberg's final chapter isn't ad hoc. It's the answer to your question.

In ordinary quantum mechanics, the wave function corresponding to some single particle state $|psirangle$, is $$psi(x)=langle x|psirangle$$ where $|xrangle$ is an eigenstate of the position operator with position $x$.

Now in relativistic QFT, with quantum field $Psi$, it is problematic to define a position operator, but $$Psi^dagger (x)|0rangle$$ creates a single particle which acts in some sense as if it is localized at $x$. It is a Fourier transform of momentum states, and it has the right transformation properties under the Poincaré group.

So if we have a single particle state $|psirangle$ in QFT, $$psi(x)=langle 0|Psi(x)|psirangle$$ is an object that is defined just like the ordinary QM wave function. And from the operator equation of motion for $Psi$ we get the wave function equation of motion for $psi$. So the solutions of the Dirac equation thought of in the wave function sense really are applicable to these states $|psirangle$ in the genuine QFT.