Energy-momentum vs charge-current Is there a simple intuitive way to explain why energy-momentum density requires a tensor, while charge-current density is a vector?
$\partial_{\mu}  J^{\mu} = 0$ is a statement, in effect, that the integral of $\rho$ over all space is time-independent: that is, charge is conserved.  In classical physics, energy is conserved.  Energy density together with energy flow density would seem to be a 4-vector very much like the flow of charge. Feynman gives a very similar argument when he introduces the concept of electromagnetic field energy and momentum density.  I'm looking for an intuitive explanation of why energy current density and charge current density can't be treated in similar ways.  
 A: 
Is there a simple intuitive way to explain why mass-momentum density requires a tensor, while charge-current density is a vector?

Yes. The four-current is the density associated with a conserved scalar, the charge. The stress energy tensor is the density associated with a conserved four-vector, the four-momentum. In each case the density is one rank higher than the corresponding conserved quantity.

I'm looking for an intuitive explanation of why energy current density and charge current density can't be treated in similar ways.

Although both are conserved, charge and energy are very different kinds of quantities. Charge is a scalar quantity (a tensor of rank 0) meaning that it is the same in all reference frames. Energy is not, different frames disagree on the value of energy. Instead, energy is a component of a vector (a tensor of tank 1).
There are very few circumstances where a scalar can be treated in a similar way as a component of a tensor.
A: Let $\Omega^{\nu_1 \cdots \nu_r}$ be some quantity of rank $r$. The density $\omega^{\nu_1 \cdots \nu_r \mu}$ is of rank $r+1$ because it will be contracted against a 3-dimensional element $dV_\mu,$ which in Minkowski space can be considered being a rank 1 tensor, e.g., with time being the first dimension, $dx \, dy \, dz \sim (dx \, dy \, dz, 0, 0, 0)$:
$$\Omega^{\nu_1 \cdots \nu_r} = \int_{\mathbb{R}^3} \omega^{\nu_1 \cdots \nu_r \mu} \, dV_\mu.$$
