First off, a Michelson interferometer does work. It can measure tiny length differences. Here is a video showing one in action.
I must have misunderstood how it works. Here's what I think is happening:
The light beam is split in two. The different paths may be different lengths. They are then recombined, and so they may interfere.
To see the interference, the light goes through a pinhole that is multiple wavelengths wide. The light spreads out in all directions, and when it hits a flat wall anywhere but the center, the light that went through different parts of the hole will have traveled different distances. So it interferes.
All that matters is the geometry. The first dark circle comes at a place where the light from the far end of the pinhole has traveled one wavelength farther than the light from the closer end. So it cancels.
Since the wavelength is the same for the two beams that were split and have come together, each of them cancels with itself at that spot. Also at the spot where the total difference is two wavelengths, and three wavelengths, etc. So those rings will always be dark. But the places where the light constructively interferes, the two beams may cancel there, or complement each other. So we can tell about tiny differences in length by how bright the unmoving bright circles are.
But that isn't what actually happens! In the video, when one mirror is moved a little, the rings actually move! They don't just get brighter or darker, they stay the same brightness but move outward, or back in.
So my understanding is somehow wrong.
What is really going on here?
I definitely misunderstood. I assumed that there was a pinhole to create an interference pattern between the beamsplitter and the screen where the interference pattern is observed. When I look more closely, no one has said this exists.
With the single pinhole or slit, I expect to get the exact same interference pattern from both beams. The big difference I see from having only one beam is that when two beams interfere the right amount we should see an interference pattern when the slit is less than one wavelength wide, but with other amounts of interference that would not be visible.
Without the pinhole or slit, the interference pattern depends on some subtle misalignment of the beams. I'm fascinated at the idea that when the lengths are equal there is no interference, and at every other pair of distances there is. @Farcher provided a link to a simulation which demonstrated this.
I'm still confused, but now I know where I went wrong, and I have an idea where to look.