Probably the bulb will burn out in a few seconds. If the battery is very low voltage, the system might operate fine though due to resistances in the bulb, the battery, and the wires.
If we attach a battery to a small light bulb without any other resistances a couple of things could happen depending on the voltage of the battery, the durability of the light bulb, and the resistances that are inherent in the light bulb, the wires, and the battery.
All parts of this system do have resistances. The internal resistance of the battery is sometimes called the Thevenin resistance. Wires have tiny resistances. Light bulbs have significant resistances but probably not enough to keep the system from breaking.
What will probably happen is that the filament on the bulb will burn out, especially if it is a small bulb which likely has a thinner filament. What is happening is that the current through the system is very very high. According to Ohm’s law, for a given voltage, current is inversely proportional to resistance, and in this case resistance is low.
If the bulb was somehow able to handle this current the next thing I would expect is for the battery to catch on fire. The high current will produce a lot of heat in the battery which can easily cause it to melt or burn.
If the battery and bulb were somehow fine, then it is possible for the wires to heat up and fail (break from thermal weakening), but I don’t see how this could ever happen before the battery or bulb failed. The last possibility, which I cannot imagine being the case unless the battery is weak, is that the system can sustain the high currents. The inherent resistances in each part are enough to modulate the current enough that the parts wont burn out. The bulb will probably be pretty bright since its light intensity would be (roughly) proportional to the dissipated across it.