The relay has a pick up voltage of 3v8 according to its spec sheet. Your supply is 5v. 3v8 is 0.76 of 5v. One time constant for capacitor charge charges a capacitor to 0.63 of the supply voltage, so you would need between 1 and 2 time constants to reach 3v8 and operate the relay.
The time constant is found by resistance times capacitance (Farads). Let's say you can tolerate a 100 mS delay in normal operation and that is enough of a delay to prevent the initial firing of the relay. (Both are guesses).
For a time constant of 80mS you could use a 100uF capacitor in series with 820R. A little over 1 time constant will then charge the capacitor to 0.76 x 5 v = 3v8 and the relay would operate.
Try a 100uF in series with 820R and wire that across the relay coil. The relay should ignore the short transient at booting time but should respond within 100mS when a real signal arrives.
I have not tried this out in practice for you. I have done the same many times before with similar circuits. Therefore "E&OE" as the lawyers say. Good luck and let me know how it goes.
If you have access to an oscilloscope, you can easily see the length of the boot time transient and re-calculate appropriately. Then you can shorten or lengthen the delay by decreasing or increasing resistance (or altering the capacitor).