Latching battery switch triggered by interrupts
What you want is a high-side PMOS switch. That is a P-channel, enhancement-mode MOSFET between the 3.7v battery and the load. Many of these exist today for precisely this purpose, capable of passing up to 50A or so. To keep it "off", tie the gate to 3.7v with a 100kΩ resistor. This will keep the gate at 3.7v normally. To turn it on, pull the gate to ground. If you suspect lots of electrical noise near this circuit and/or want faster turn-off, decrease that resistance value. There is a lot more to this, but that is the basics. 

If the interrupt pin is an open-collector type, and is active-low, then tying that to the gate directly should do it. Low = on, open = 3.7v. Verify with the device datasheet that the pin can handle 3.7v.

If it is not open-collector, and the digital logic voltage is less than 3.7v (such as 3.3v or 1.8v), then that is an issue. Most pins have clamping diodes on them internally, and raising a pin above Vdd always has bad consequences. In that case, just use an NPN transistor to sink the gate current. That "isolates" the 3.7v from the pin, which now behaves as High = on and Low = off.



Note that the IRF7406 is a 3A device. If you need more current than this, then a different MOSFET will be required. Likewise, if you only need 100mA, then a physically much smaller one can be used instead. Also note that MOSFETs are very sensitive to ESD. Safe handling precautions are required.