Yes. Field Weakening as it is known, is a tried and tested method of increasing the speed of wound field motors over a relatively small range, from the days when electronics for speed control didn't exist. It is usually done by altering the current in the field coils, rather than physically moving permanent magnets around. It is responsible for the steep speed/torque curve in series wound motors still used today.
Unfortunately, the increase in speed comes not only at the expected cost of reduced torque, but also at the cost of reduced motor efficiency.
You would expect a reduced torque. The output power is speed.torque, and altering the field is not going to change the power, only the ratio of speed and torque.
You expect 'something' to be going wrong with reduced field, after all, we work hard in motors to get a good magnetic circuit, minimise air gaps, and nobody uses a motor with zero field (field weakening taken to the extreme), so what's going on?
The problem is the DC resistance of the copper windings. As you reduce the field, you need proportionally more current to maintain the torque. However, the heating losses rise as the current squared. If the field in a motor is increased (by the use of high field magnets as in modern BLDC motors), it allows the number of turns on the rotor to be reduced, allowing thicker wire, which reduces copper losses.
Let's put some numerical meat on those hand-waving bones. Let's say you have a 100W output motor, delivering 100 rad/s and 1Nm. The rotor is wound with 2 equal windings of 0.2ohm each, which we can connect in either series or parallel. The magnetic field is such that the motor draws 10A when producing its 1Nm torque, with the rotor windings in series. Ignore all other losses.
The rotor resistance is 0.4ohms, dropping 4v at 10A current. The back EMF is 10V (motor with no other losses, 100W/10A), so the motor input voltage is 14v under these conditions, giving about 70% efficiency.
Now let's halve the magnetic field, keeping the current the same. The output torque is halved to 0.5Nm. Let's assume we load the motor such that the speed can rise to 200rad/s. With half the field and twice the speed, the back EMF is the same at 10v. With the same current, we have the same voltage drop, the same 14v supply, and the same efficiency of about 70%. So what's happening here with my assertion that the efficiency falls?
Let's consider the other way of doubling the speed of the motor, keeping the magnetic field the same, and putting the rotor windings in parallel. This halves the effective number of turns, which halves the back EMF, and halves the output torque. If we load the motor so that the speed again can rise to 200rad/s, the back EMF is now again 10V, and we need 10A to get the 0.5Nm output torque. However, the rotor now has a resistance of 0.1ohm instead of 0.4 ohm, so the IR volt drop is 1v, meaning a motor supply voltage of 11v. Our efficiency has increased to about 90%!
So now what's going on, does this mean that increasing the speed of the motor keeping the field the same increases efficiency? What about increasing it to very high speed indeed? We have ignored other losses, like air resistance, and the increased eddy current losses in the rotor iron, and there are other costs of high speed, like the need for rotor strength, and the need to balance it, and the problems of reversing the field quickly in the windings. So we cannot improve efficiency ad infinitum by increasing speed, there will be an optimum.
The take-home is that the highest field possible gives the best efficiency, and the rotor should have the correct number of turns to match the desired speed. However, reducing the field in a wound stator motor is a neat way of controlling the speed, at the cost of efficiency.