How to provide stable power to RF receiver when operating on small batteries with blinking LEDs
An adequately good regulator and/or just designing the circuit properly will very probably be cheaper, more compact and less complicated than using additional batteries. 

Showing us your "very simple circuit" is essential.
 It is very likely from what you say that you have made it simpler than possible - something that Albert Einstein warned against doing.

Please show us your circuit.
Without a circuit and full information your question can be summarised as "Please tell me."

What is your design LED current?
How often does the L:Ed flash?
How long is the LED on for ?
What is your superregen maxmum current drain?
 hat is the minimum voltage that your superregen will operate on?  

An AG13 / LR44 / ... "coin cell" is not designed for currents more than a few mA and ideally should be operated at less than 1 mA. Internal resistance is probably in the 2 to 10 ohms range, and peaks caused by an LED flashing, if taken directly from the battery, will cause the actual battery voltage to fluctuate significantly.
 At a minimum you want a good capacitor connected across the battery. I'd guesstimate that 100 uF would be a minimum and more would do no harm. 

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Try this:

(1) Supply your super-regen via a resistor of value
 Rsr = 500/Isr ohms,
 where Isr is the average super-regen supply current in mA. eg if the SR can draw 5mA max then Rsr = 500/5 = 100 ohms.
(2) Connect as large a capacitor as you reasonably can from the super-regen supply on the SR side of Rsr to ground 1000 uF would be good. More would be better. 10,000 Uf would be good for experimenting. If it works it shows us what to focus on. These Pansonic 10,000 uF, 6V3 capacitors cost about $1 in 1'2 from Digikey  -use one as above and then if that works we can work on a better and cheaper solution. 

But - an adequately good regulator will probably be cheaper, more compact and less complicated than using additional batteries. )



How large a capacitor?

When a battery is unable to to supply a stable voltage during a sudfden load peak, adding a capacitor "across the battery" my allow the voltage "blip" to be reduced to an acceptable level. Or not. 

In my above advice the suggested 10,000 uF may sound immensely too large.
 Is it?

If you can afford to run the receiver on for less voltage that the battery provides then a logical thing to do is to provide a filter capacitor that feeds a regulator. The output of the regulator is set to as low a voltage as can be tolerated by the receiver  and this allows the input capacitor o be at battery voltage before the LED "blip" and to drop to not quit so low as to have the regulator "drop out" during the blip. 

BUT my above solution eliminated the regulator. The aim was to greatly reduce the battery voltage dip by having a very large reservoir capacitor. 

This is not an approach I would usually favour, but it matched the user's question and will show whether it is a practical solution.

But, how big a capacitor is needed. 
 A short heavy load such as a single  LED flash will require a known amount of energy. The effect of taking this amount of energy from a capacitor can be calculated.
 Here's an example.
 Imagine that the LED will operate at 2V at 20 mA for 0.1S.
 Decide that a 0.1V drop in voltage is OK. 
 The energy in a capacitor of capacitance C at Voltage V is given by 0.5 x C x V^2.
 If the Voltage falls from Vi (Vinitial) to Vf (Vfinal) the energy taken is 0.5 x C x (Vi^2 - Vf^2)
 The energy in the above LED fash = V x I x t Joule. We use full bttery V as the LED uses 2V BUT the extra voltage drops across the LD drive resistor. Decide to try this for Vi - 3V.

So Vi x Iled x t = 0.5 x C x (Vi^2-Vf^2)  so
 C = 2 x Vi x Iled x t / (Vi^2 - Vf^2)
 = 2 x 3 x 0.020 x 0.1 / (3^2 - 2.9^2) = 0.0203 F =~ 20,000 uF.
 Close enough to the 10,000 uF = 10 mF that I suggested.
 Using a 0.2V blip will allow 10,000uF to be used.

[mumbo jumbo]    &lt;- formatting language dos not like triangular brackets here.
  A sobering lesson occurs if we start allowing larger and larger dips in voltage.
At first the needed cap size falls as the allowable dip falls. 
 BUT after we allow a dip of about 0.5V and needed cap = 4000 uF we notice that the cap size does not fall much more as the allowable dip continues to  increase.
 At 1V dip we need 2400 uF and at 3V dip (capacitor is fullt drained by the LED we still need about 1300 uF.
 The reason is that the V^2 term in the equatin means that most of the enrgy is stored at the top of the voltage decline with the balance left over as voltage falls decreasing as the inverse square of the voltage left. [Dont worrry about it]
 As the LED stops drawing current below about 2V such large dips are in any case unrealistic - our 0.1 to 0.2V is more appropriate and at best the capacitor can fall to Vled.
 [/mumbo jumbo]

So, yes, 10,000 uF is  not a terrible choice, and if we want to use a regulator with Vout = say 2.2V then we are still going to need a cap most of this value.