Please allow me to express my views in an attempt to further simplify this explanation. Experts are welcome to suggest any improvements.
In the scientific representation mentioned by the OP, Mantissa M is represented by 9 bits (0-8) and Exponent e by 6 bits.
Hence, the maximum value that can be expressed by Mantissa and Exponent bits are 511 and 63 respectively. However, as (in IEE 754), Exponent bits all 1s and Mantissa non-zeros means NaN, so the max possible value for E bits shall be 62.
Now the story starts as:
Minimum positive number being -> Minimum M and minimum E.
Hence, in scientific notation, [1 + .000000000] x 2^(e-bias) = 1.000000000 x 2^-31
which can be written as [1 x 2^-31 + (0 x 2^-9 x 2^-31)]. This is the smallest positive number which can be represented.
The next number in sequence will be (increment that 0) (2^-31 + 1 x 2^-9 x 2^-31)
This can go on till the increment reaches 511.
The next increment will be possible only by incrementing the exponent and resetting the Mantissa to 0.
(Ex: 1.8 -> 1.9 -> 2.0)
This way the maximum exponent that can be reached is 2^(62-31) = 2^31.
With this, the Mantissa increments will be (2^31 + [0 to 511] x 2^-9 x 2^31) which implies that every single increment from [0 to 511] causes a jump of 2^22.
It is also worth noting that the general maximum possible number using the above notation is (2-2^-9) x 2^31. It is 2-2^-9 because adding one more in the 9 decimal bits (1.111111111) will make it 2.
Hope this helps others too!