Your bipolar 555 draws several milliamps when idle.
CMOS circuits of the same complexity usually draw very low leakage currents (<1µA) when idle, and only draw significant currents when switching.
Switching in CMOS is similar to a capacitor charge/discharge. The amount of charge needed is proportional to supply voltage. Therefore, average current draw is proportional to Vsupply multiplied by switching frequency.
It is thus very desirable to lower the supply voltage. If you can use a chip that will run on an unregulated 1.8 ... 3.3V, then it will run on 2 AA batteries in series, which will take up about the same space as 9V battery.
However, the two AAs have about 4x the capacity in mAh, and lowering the voltage from 9V to 3V divides current consumption by 3x. This increases your battery life at least by 10x, easily. If you can do the rest of the circuit with 2 AAs, then do it. Also, 2AA are much cheaper than one 9V!
You can use CMOS like the CMOS 555, 4000 series, CMOS 74 series, or a modern microcontroller. Simply check that they are compatible with your supply voltage.
For example, 74LVC and ALVC operate From 1.65 V to 3.6 V. Many modern microcontrollers will also run on 2 AAs until they're thoroughly exhausted... even good old Atmega328P runs from 1.8 to 5.5, altough it's not an ultra low power device.
Using a DC-DC converter (switching or linear) is not always a good idea, as its idle dissipation can be much higher than what the circuit actually uses, unless it is designed for this specific purpose.
You can also run your 1.8V device from 2 AA using a micropower 1.8V LDO, which will lower voltage, and thus current draw, even more.
However... we don't know if your circuit will have some power-hungry loads, so more details would be needed.