The Dirac equation can be defined as
$$D\psi = \Sigma_{i=1}^3 c(e_i) \nabla_{e_i},$$
where
$$\nabla_{e_i} = e_i + \frac{1}{4} \Sigma_{j,l=1}^3 \Gamma^l_{ij} c(e_j) c(e_l).$$
$c$ is the standard representation of the Clifford algebra on the spinor bundle and $\Gamma^l_{ij}$ are the connection coefficients for the metric $g$. In particular, a spinor $\psi$ is a harmonic spinor if it satifies
$$D \psi =0.$$
If for some reason, the Christoffel symbols were vanishing, one would have for the harmonic spinor equation
$$(c(e_1)e_1 + c(e_2) e_2 + c(e_3) e_3) \psi =0.$$
Does this equation have existence of non-zero solutions because one can choose orthonormal frame fields $e_i$ for the metric $g$ such that it holds? It seems by inspection that this equation cannot have non-zero solutions because you will just have a constant times $\psi$ equals zero, which only holds for the zero spinor. If one could have the coefficient equal to zero by a choice of frame fields, then any spinor would be harmonic and solve the equation.