Detailing why a scalar gravity theory predicts no bending of light [closed]

I want to understand in technical detail why a particular scalar theory for gravity predicts no bending of light. It is left as a question, either in "Gravitation" by Misner, Thorne, and Wheeler, question 7.1, parts A and E, or in a paper by Thirring (Ann. Phys., 16:96-117, 1961). I am not in graduate school for physics (and so this is not a homework problem for me, just self-study). I am getting stuck at the end of 1.A so cannot proceed to 1.E. Here is the problem

Exercise 7.1. Scalar Gravitational Field, $\Phi$

A. Consider the variation principle $\delta I = 0$, where $$I = - m \int e^{\phi} \sqrt{ -\eta_{\alpha \beta}\frac{d z^{\alpha}}{d \lambda}\frac{d z^{\beta}}{d \lambda} }~d \lambda,$$ Here $m = (\rm{rest~mass})$ and $\lambda = (\rm{parametrized~world~line})$ for a test particle in the scalar gravitational field $\Phi$. By varying the particle's world line, derive differential equations governing the particle's motion. Write them using the particle's proper time as the path parameter, $$d \tau = \sqrt{-\eta_{\alpha \beta}\frac{d z^{\alpha}}{d \lambda}\frac{d z^{\beta}}{d \lambda}}~d \lambda,$$ so that $u^{\alpha}=d z^{\alpha}/d \tau$ satisfies $u^{\alpha}u^{\beta}\eta_{\alpha \beta}=-1$.

The scalar gravity field $e^{\phi}$ is along for the ride in part A. It is simple enough to take 4 derivates of the action with respect to the 4 parametrized velocities $d z^{\alpha}/d \lambda$. I remain confused as to how to "[Re?]write them using the particle's proper time as the path parameter".

For completeness, I will include part E which will be my next hurdle after A is understood.

[B, C, and D involve the field equations and perihelion precision.]

E. Pass to the limit of a zero rest-mass particle in the equations of motion of part A. Do this by using a parameter $\lambda$ different from proper time, so chosen that $k^{\mu} = d x^{\mu}/d \lambda$ is the energy-momentum vector, and by taking the limit $m \rightarrow 0$ with $k^0 = \gamma m = E$ remaining finite (so $u^0 = \gamma \rightarrow \infty$). Use these equations to show that the quantities $q^{\mu} = k^{\mu} e^{\Phi}$ are constants of motion, and from this deduce that there is no bending of light by the Sun in this scalar theory.

1. Start with the action: $$I = - m \int e^{\phi} \sqrt{ -\eta_{\alpha \beta}\frac{d z^{\alpha}}{d \lambda}\frac{d z^{\beta}}{d \lambda} }~d \lambda$$
2. Replace $\lambda$ with $\tau$ $$I = - m \int e^{\phi} \sqrt{ -\eta_{\alpha \beta}\frac{d z^{\alpha}}{d \tau}\frac{d z^{\beta}}{d \tau} }~d \tau$$
3. Write the Lagrangian as: $$\mathcal{L} = - m e^{\Phi} \sqrt{-u^{\mu} u_{\mu}}$$
4. Use Euler-Lagrange to calculate the equations of motion: $$0 = m e^{\Phi} \frac{d}{d \tau} \frac{u_{\mu}}{\sqrt{-u^{\nu} u_{\nu}}}$$ Because we are working with a worldline, the contraction of $u^{\mu}$ will create the ratio of an interval over an interval, good old unity. To dodge the imatinary bullet, it must be negative unity. I am not clear about the two minus signs in the action, but I will not lose sleep over it.

Start by varying the action in the case where $\phi=0$, i.e., no gravity, and learn how to derive the equations of motion for a particle of mass $m$. You should be in a better position to approach this problem then.