The reason different answers disagree is that your question is underspecified. You need to say more about how the water moves.
Scenario 1: steady state flow
Suppose you had a long pipe with water flowing at a constant speed $v$. How much energy is required to keep the water flowing at that speed? Of course, in the absence of viscous or turbulent losses (which I will neglect throughout this answer), the answer is just zero, just like how it takes zero energy to keep a block sliding at constant speed on a frictionless surface.
So to get a nontrivial answer, you need to specify how new water is brought from rest to speed $v$. That is, you need to think about how water enters the pipe.
Scenario 2: conservative flow
Suppose we stick the end of the pipe in a container full of water. We either run a pump to get the water to go through the pipe at the desired speed, or if the water container is high enough, we can just use the preexisting hydrostatic pressure. We will need to spend energy to get water to flow through the pipe, either as electrical energy in the pump or gravitational potential energy in the water in the container. For concreteness, let's say the pipe is attached near the top of the container, so the energy just comes from a pump.
If there are no viscous or turbulent losses, then the power that you need to input is equal to the rate of change of kinetic energy of the water,
$$P = \frac{dK}{dt} = \frac12 \frac{dm}{dt} v^2 = \frac12 \rho A v^3.$$
This is the result given by several of the other answers.
Why doesn't your argument work?
It's somewhat subtle. First, as the other answers have said, if you put a pump at the entrance to the pipe, then it speeds up the water as it passes through. Therefore, you can't use the equation
$$P_{\mathrm{pump}} = F_{\mathrm{pump}} v$$
because the speed of the water is actually continuously increasing from zero to $v$. However, this isn't a complete answer, because you could also place the water pump in the middle of the pipe, where the water is already flowing with a uniform speed $v$. In that case, assuming an ideal pump, your equation $P_{\mathrm{pump}} = F_{\mathrm{pump}} v$ is perfectly correct.
Here the problem is actually your first equation, which implicitly assumes that the pump's force is the only external force in the system. Think about the entrance to the pipe. The water outside it is moving at almost zero speed, while the water inside is moving at substantial speed. Thus, by Bernoulli's principle, the pressure is higher outside the pump, and this pressure difference provides an additional force that helps accelerate the water. In fact, for certain idealized geometries, it is actually exactly equal to $F_{\mathrm{pump}}$, which implies that $F_{\mathrm{pump}}$ is only half the net force. Inserting this factor of $2$ recovers the correct answer.
This factor of $2$ is a well-known effect in plumbing (i.e. applied fluid dynamics), which leads to the phenomenon of vena contracta. It generally isn't mentioned in introductory physics, because it's somewhat subtle, but the pioneers of fluid dynamics always accounted for it, as otherwise all their results would be totally wrong.
Scenario 3: inherently inelastic process
There are also situations where something like your answer is correct. For example, suppose you had a very long circular trough, through which water runs at speed $v$. Now suppose that rain falls onto the trough, increasing the water mass at rate $dm/dt$, while a pump inside the water keeps it going at constant speed. Then your derivation would be correct, because there are no important external forces besides the pump's, so
$$P = \frac{dm}{dt} \, v^2$$
with no factor of $1/2$. On the other hand, by energy conservation we have
$$P = \frac12 \frac{dm}{dt} \, v^2 + \frac{dE_{\mathrm{int}}}{dt}$$
where $E_{\mathrm{int}}$ is the internal energy of the water. These two expressions are compatible if
$$\frac{d E_{\mathrm{int}}}{dt} = \frac12 \frac{dm}{dt} \, v^2.$$
But didn't I say earlier that we were ignoring all viscous and turbulent losses? Well, in this case, you can't. If there was no viscosity or turbulence, then the rain water would just sit still on top of the flowing water, and the pump would require no power at all. Some kind of dissipation is required to get the rain up to speed $v$. Since this is essentially an inelastic collision, it dissipates energy, making the pump power twice as big as the naive result. This is another famous factor of $2$ (unrelated to the previous one) which does show up regularly on tricky high school physics problems.