Your conclusion that the first motor is the most powerful is correct, given those figures. Note that its peak current is 5A, so you'll need quite a substantial motor driver and battery; either a 4-cell LiIon pack with a suitable rating or a car battery. You may find a car battery a useful weight for holding the whole system on the ground!
Before looking at motors, you need to define a few parameters, apart from the load mass and leadscrew pitch. First, what is the motion profile? I assume the device won't be in constant motion. Is there a defined sequence of movements? What distance and elapsed time for each move? What elapsed time between moves?
Next, the load. Is it affected by gravity? Are there other external forces acting on the load? What about friction? You should also know the diameter and material of the leadscrew (for inertia calculation).
With all of that, you can proceed to calculate peak torque, RMS torque, and speed required at the drive end of the lead screw. That will indicate what sort of motor you need.
Then you need to consider the ratio of the reflected load inertia to the motor's rotor inertia. A high ratio will necessitate de-tuning the position loop in order to get stability. Ideally, the ratio is 1:1 (hard to achieve), but under 3:1 is generally very good. Under 6:1 is generally acceptable. If the coupling is "stiff", 10:1 can be OK.
If the inertia ratio needs reduction, first look at a bigger motor. Otherwise, a gearbox may be required. Reflected inertia reduces as the square of the gear ratio. But speed requirements increase.
Best Answer
No load current is zero amperes.
KV 2100 = 2100 RPM/V = 35 R/Vs = 220 rad/Vs
So, Ke,Kt = 1/220 = 0.00455 [Vs/rad] = 0.00455 [Nm/A]
$$U=K_e\cdot\omega + R\cdot I $$
EDIT:
\$T=K_e\cdot I \$ or \$I=\dfrac{T}{K_e}\$ so if T=0 then current is zero. Else you would have to know the no load torque, which could be the ball bearing friction and air drag at known speed.
The derived Ke from KV that was actually measured in laboratory consists of a test setup. This is a second motor coupled to the shaft and spin as much as the current is zero - no load current. In such situation the applied voltage equals the back emf voltage \$ U=K_e\cdot\omega\$, knowing \$U\$ and \$\omega\$ is it possible to estimate \$K_e\$ or KV in your case. So the motor is spinning at 35,500 RPM at 15V, but not by himself, rather by the help of the coupled motor from the setup.