Why is the MOSFET driver in this circuit dying?

One obvious difference between a resistive load and a battery is that on startup, the resistive load is zero and the battery obviously isn't - it's a source after all. This is called a "precharged load" and can be messy without some form of active or passive OR-ing control.

You don't have any isolation between your synchronous buck and the battery, so if your PWM commands low duty cycle, your "T2" (it really should be Q2 or U2, since T is generally used for transformers) will be on with a wide duty cycle and sink current through the inductor. Your synchronous buck may start acting like a synchronous boost, or even saturate out and short the battery. As "Some Hardware Guy" stated, when the battery is connected you may see some interesting things on pin 6.

Also, there are some intelligent buck controllers out there that can sense the low-side switch drop and shut things off if heavy current is detected. The IR2104 is a high-side / low-side half bridge driver, which can be used to drive a buck, sure, but it's not the best choice.

Try a big diode between your buck and battery, and see if the behaviour changes.

Application note AN-978 has some notes on sizing the bootstrap capacitor C3 which may be relevant. The examples use a considerably larger value than 0.1uf

Madmanguruman and David Tweed are probably on the right track here.

Here is a scenario:

When the circuit starts the synchronous FET (T2) will be on. It has to be this way because there is no bias voltage for the high side drive for T1 except through T2. So, T2 sinks current from the battery through the inductor. Then finally T2 turns off. It doesn't matter if T1 turns on or not, the inductor dumps current back through T1 (or its body diode). The current has to go some place. SMPS are 1 quadrant devices (almost always), so the Laptop supply sinks nothing. The current can only go into C5 and Vcc of the IR2104, which has a max supply voltage rating of 25V. Boom.