If the number in question is known to be a power of 2, you are just trying to find $n$ such that $N=2^n$. If the input number is in decimal notation, just count the digits, subtract 1, multiply by 3.3219, and round up. Then add 1 if the initial digit is 2 or 3, 2 if the initial digit is 4, 5, 6, or 7, and 3 if the initial digit is 8 or 9.
For example, suppose $N=1267650600228229401496703205376$. This has 31 digits; $30\cdot3.3219 = 99.657$, so $N=2^{100}$. Or take $N=$
43699499387321412970609716695670835099367888141129535719972915195176
79444176163354392285807163181819981286546206512408458617685052043667
09906692902245553277900892247131030458103436298545516643924637451297
481464347472084863384057367177715867713536
which has 246 digits. $245\cdot3.3219 = 813.872383$; we round up to 814, add 2 because the first digit is 4, so this number is $2^{816}$.
The magic constant 3.3219 is actually $\log 10 / \log 2$. For input numbers in the hundreds of thousands of digits you will need a more accurate version, say 3.3219281.
Yes, your proof is correct. You have shown that there are no odd perfect numbers that are squares.
The conjecture that there are no odd perfect numbers remains open.
Best Answer
If you already know the number is a power of 2, then all the factors are also powers of 2. So, if $n=2^k$, then the factors are $1, 2, 2^2, \dots 2^k$, and there are exactly $k+1$ of them.