$$\mathcal{L}=\bar\psi(i\gamma^\mu\partial_\mu-m)\psi,$$ taking Euler-Lagrange equation on $\bar\psi$ gives the more familiar Dirac equation $$(i\gamma^\mu\partial_\mu-m)\psi=0$$ and its adjoint version $$\bar\psi(i\gamma^\mu\stackrel{\leftarrow}{\partial_\mu}+m)=0 .$$
Taking E-L equation on $\psi$ however gives $$\bar\psi(i\gamma^\mu\partial_\mu-m)=\partial_\mu(\bar\psi i\gamma^\mu)$$ which clearly constrains $\bar\psi$ differently. What is going on here?
(I haven’t checked the equations of motion you have written, I am assuming they are right)
In quantum field theory fields $(\phi)$are operators but positions $x$ are not. So $\partial_\mu$ is also not an operator. That means conjugate of $ {\partial_\mu\psi}$ is just $ {\partial_\mu\bar\psi}$
So the adjoint of EOM with respect to $\psi$ is actually $$(i\gamma^\mu\partial_\mu+m) \bar\psi =0 $$
Now use product rule on EL equation on $\psi$ to see that they are the same.
