You have a positive-feedback oscillator, because the +Vin (pin2) is routed directly under the Vout (pin7).
EDIT Also, the Rfeedback 604Kohm drives 5pf or more, what with 3 causes of capacity on Vin+ :::
1) the input capacity of the comparator
2) the side capacitance of that long Vin+ trace to the surround GND
3) the many bits of metal over that long Vin+ trace
EDIT The 604kohm is too large to quickly implement 15Kohm/600Kohm * 5v = 150 millivolt of hysteresis. Reduce your resistors on Vin+ by 10X, to 1.87K, 10.7K, 60.4K
Assume you have 2mm by 4mm area of coupling between pin7 and pin2, on pin7 side of the IC. With PCB thickness of 1/16". What is the capacity?
C = Eo * Er * area/distance = [9e-12Farad/meter * 5 * 4mm * 2mm / 1.5mm] * meter/1000mm
C = 9e-12 * 5 * 6/1.5 * 1/1000 = 180e-12-3 = 180e-15 = 0.18pF
Assume Cnode(pin2) is 3.6pF. The 5 volts output swing is attenuated by 0.18/3.6 or 1/20 thus V_injected_pin2 of 5/20 = 0.25 volts.
Place a scope probe on Pin2, and see that 0.25 volts (but will be further attenuated by 10pF or 15pF of the scope probe, to approximately 50 milliVolts).
The timeconstant on Vin+ is 0.18pF * 15Kohm or 3nanosecond feedback thru the parasitic cap I computed. There is also parasitic across the feedback R (604Kohm).
Cure? solder-tack a 10pF across the 604Kohm Rfeedback, to ensure the hysteresis is LARGE, and slowly decays. Use a 10pF leaded, to make it easy.