How does the power requirement change when weight and speed change?
Short answer: No. $ P = F \cdot v$ and since $ D = \frac{1}{2}C_D\rho v^2S $ it follows that $P\propto v^3$

Let's take a more simple case: a block moves over a rough, horizontal surface in a vacuum and a force is applied such that its velocity $v$ stays constant. This force is always the same, since the "drag force" is given by the friction coefficient times the normal force. Here, the kind of relationship that you're thinking of would hold: if we're moving at $2v$, we need twice as much power ($P=F\cdot v$). 

Look at a fixed distance along that trajectory: we do the same work in both cases ($v$ and $2v$), so we put in the same energy which dissipates through friction, i.e. heat. However, we now need to do the same work in half the time, so the power is double. In a certain way, it's "more difficult".

Now for an aircraft, the drag force itself is proportional to the square of the velocity. So, in conclusion, there are "two effects" which "add up": a quadrupling related to the increased force and a doubling due to the increase in velocity. Hence, the power is multiplied by 8.



Edit (in response to comments below):

Ok, let's do another thought experiment: we place a jet engine in a wind tunnel and keep it at constant thrust and constant mass flow (choked nozzle), which results in a constant velocity increment across the engine. The power contained in the jet (the airflow exiting the engine at higher speed $v_j$) is $\frac{1}{2} \dot m v_j^2$  and the power contained in the incoming flow is $\frac{1}{2} \dot m v_0^2$. The difference between the two is called the "propulsive jet power"; it's the power which is "available" from the acceleration of the flow. Now, you can see that because both terms are squared, it is not just the velocity increment which counts (as it was the case for the thrust), but also the magnitude of the two terms.

So alternatively to the "thrust power" given by
$$P_{thrust} = \dot m (v_j-v_0) v_0 = T v_0 = D v_0$$

we can also consider the "propulsive jet power" 

$$P_{prop} = \frac{1}{2} \dot m (v_j^2-v_0^2)$$

where the proportionality to $v_0$ at constant thrust is maybe more clear. The difference between these two quantities is called the power loss ($P_{loss}=\frac{1}{2} \dot m (v_j-v_0)^2$) and is a measure of how much of the orginal chemical energy contained in the fuel is lost to kinetic energy in the jet. 

You can see that if the velocity differential is constant, the power loss is constant. So if we have shown that the propulsive jet power is proportional to $v_0$ at constant thrust, the thrust power (which is the power that "actually moves you") must also be proportional in this way. And this proportionality is related to the fact that the kinetic energy of a mass is proportional to the square of its velocity.