You can use L+C loops, resonated at some frequency, and trigger the ringing with a sharp edge from some logic device/gate. Simply connect the gate to series LC circuit to Ground. That is the transmitter. For each "1", send an UP_DOWN pulse to the LC resonator. For "0", do nothing. You may need a bunch of start bits, so the receiver knows it needs to pay attention.
For the receiver, use an identical L+C loop, similar size, so resonates at the same frequency. Simply connect the L+C loop to a PeakDetector. For a very simple decoder, clock the PeakDetector output into a 8-bit ShiftRegister (74HC595). Assume 2 start bits (to wake up the Receiver shifting bit-rate clock). You'll also need to define a packet length, and have enough 8-bit SR to clock in that entire packet.
Now. The science part. How much power must you transmit, across 6 inches?
Turns out the necessary receiver power depends on your bit-rate and your bit-error rate. Lets compute that minimum receiver power.
To avoid a bunch of multiplications, we'll use the standard method for communication links ----- logarithms, which engineers have chosen to scale up by 10 [for more resolution], and name deciBels to honor Alexander Bell. Here goes.
-174 .......... dBm/rootHertz.....this is Boltzmann's Constant * temperature
+60 .......... dB for a bandwidth of 10^6 cycles/second or 1MegaHertz; this statement of system bandwidth also helps, by cancelling the 'rootHertz'
+20 .......... dB for how much stronger we need the signal to be, compared to
the Boltzmann noise; here the signal is 10^(20/10) stronger, or 100X
By adding these, we know how strong a signal, how much electrical energy, we must get from the Receiver antenna: in our case, that L+C loop.
Lets add -174 +60 +20 == -174 +80 == -94dBm. (no multiplication needed)
What next? Now convert from -94dBm to milliWatts. Engineers have defined 0dBm as 1 milliWatt. Thus +10dBm is 10milliWatt. And -10dBm is 0.1 milliWatt. Our -94dBm is -100dBm + 6dB. The -100dBm is 10^-10 weaker than 1mW; the +6dB is 2 factors of 2, or 4.
Our necessary receiver power, -94dBm, is 4 * 10^-10 milliWatts. Or 4 * 10^-13 watts.
What power can we get from a Transmitter, from a logic-gate? Using $$Power = Voltrms^2/Resistance$$, with 0.223 voltsRMS from a logic-gate, into 50 ohm resistor, we get exactly 1 milliWatt, or 0dBm. Peak to Peak, that's 0.632 volts.
Any logic gate can provide that.
Your transmitter is just a logic-gate into a resonant LC series circuit to GND.
Your receiver uses identical LC resonant series circuit, into a Peak Detector, into a ShiftRegister.
If your inductors (the "L") are 2" across, with TX and RX inductors facing each other only 2" or 3" or 6" apart, even given distance^3 attenuation.....now you need to find the math for wireless power transfer. Even with 30dB attenuation, you have plenty of communication link margin.
To reduce power into the TX resonator, use voltage divider: 100 ohms into 10 ohms, or 100 ohms into 1 ohm. Or 1,000 ohms into 1 ohm.