why do we look at dark energy as an energy? We call dark matter dark matter, because of the way it acts on galaxies, but why do we look at dark energy as an energy? With $E = m c^2$, can't dark energy be mass as well? Why dont we look at dark energy as a mass, in the same way as we look at dark matter as being a missing mass (as halo around galaxies)?
Since relativistic physics almost always takes units of $c = 1$, there is no difference between units of energy density and mass density. Conceptually, however, the gravitational charge is energy, not mass.

In general relativity, a perfect fluid has no viscuous shear stresses and no heat flux, and can be described by the stress-energy tensor $T^{\mu\nu} = (\rho + p)u^\mu u^\nu + pg^{\mu\nu}$, where $\rho$ is density, $p$ is pressure, and $u^\mu$ is the fluid four-velocity. In the case of vanishing pressure, the result is called a dust, and is appropriate for modeling cold dark matter, galaxies on the cosmological scale, etc. On the other hand, a radiation fluid has $\rho = 3p$, and is appropriate for modeling incoherent electromagnetic radiation, or a fluid made of strongly relativistic particles (e.g., hot dark matter), and so forth.

Thus, in the specific context of cosmology, the matter-dominated regime has $p\sim 0$ and radiation-dominated regime has $p\sim\rho/3$. Note that in the latter case, the fluid particles have zero (or at least negligible) mass. In-between, we can talk about a fluid of particles that are not strongly relativistic, and hence have some mass.

On the other hand, the standard model of dark energy is a cosmological constant, which is equivalent to a perfect fluid $p = -\rho$. It does not fit into the above scale; it is not matter in the above sense, nor is radiation, nor anything in-between. Furthermore, it has the peculiar property of being exactly the same in every local inertial frame. This frame-independence enables us to think of the cosmological constant as an intrinsic energy density of spacetime, but it also makes it inappropriate to think of it as a collection of fluid particles that have some rest frame.

For your purposes, this important because $E = mc^2$ is only valid in a rest frame.