In a curved spacetime, the wave four-vector $k^a$ for a solution to Maxwell's equation (in the appropriate geometric-optics limit) satisfies the geodesic equation:
$$
(k^a \nabla_a )k^b = 0.
$$
Effectively, this means that "the derivative of $k^a$ in the direction $k^a$ is zero". (Note that the first term is just a directional derivative in the direction of $k^a$. What's more, it is always possible to set down a set of coordinates in a "tube" surrounding the worldline of this geodesic, called Fermi normal coordinates, such that all the Christoffel symbols vanish at points along the geodesic. (See here for a reference to how this is done for null geodesics.) In this case, if $\lambda$ is our coordinate along the null geodesic, then the geodesic equation as expressed in these coordinates is just
$$
\frac{d k^\mu}{d\lambda} = 0.
$$
In other words, the components of the wave vector are constant in these coordinates. So in this sense, the energy & momentum of the photon is conserved in the "photon frame" (whatever that means.)
But of course, we're not photons, and so we observe gravitational redshift. In a lot of cases, though, we have a stationary spacetime, in which (roughly speaking) there's a time coordinate $t$ that the metric doesn't depend on. Formally speaking, we want there to be a vector field $t^a$, pointing in the direction of increasing $t$, such that $\nabla_a t_b + \nabla_b t_a = 0$. In this case, $t^a$ is called a Killing vector field, and it's not too hard to show in such an instance that
$$
k_a t^a = \text{const.}
$$
In this sense, a quantity related to the energy of the photon is conserved as the photon travels. For example, if we're just dealing with flat spacetime, then $t^\mu = (1,0,0,0)$ and we have $k^t = \text{const.}$, i.e., the energy of photons is constant for observers whose four-velocity points in the $t$-direction. More generally, in a static spacetime, we'll have something like $k_t = k^t g_{tt} = \text{const.}$ instead. Similarly, if we have a spacelike Killing vector field (for example, $x^a = (0,1,0,0)$ in flat spacetime), then we can define a conservation law for particles that travel along geodesics which reduces to conservation of momentum in the case of flat spacetime.
Finally, it's important to note that the actual energies & momenta that an observer will measure are related to the observer's four-velocity $u^a$ and their set of spatial basis vectors. For example, we have $\omega = u^a k_a$. Since $u^a u^b g_{ab} = -1$ by definition, this means that in general an observer whose four-velocity is in the $t$-direction will have $u^\mu = ((-g_{tt})^{-1/2},0,0,0)$, and such an observer will then see the photon as having
$$
\omega = k_t u^t = \frac{\text{const.}}{\sqrt{-g_{tt}}}.
$$
In this interpretation, the gravitational redshift is in fact solely due to the clocks of various observers "running slow" due to the $tt$-component of the metric being different. However, it's important to note that the conserved quantity $k_t$ is not necessarily an energy as seen by any particular observer (even the observer who emitted it in the first place.)
A good reference for this business with Killing vectors and conserved quantities is Hartle's Gravity: an Introduction to Einstein's General Relativity. Killing vectors are explained in Section 8.2.
