Sum or difference of two powers of two

Python 2, 43 bytes

lambda n:[len(bin((n&-n)+k))-3for k in n,0]

Say that n==2^a ± 2^b with a>b. Then, the greatest power-of-2 factor of n is 2^b, and we can find it using the bit-trick 2^b = n&-n. That lets us compute 2^b + n, which equals either 2^a + 2 * 2^b or just 2^a. Either one has the same-length bit-length as a*. So, we output the bit-lengths of n&-n and (n&-n)+n, computed from the lengths of their binary representations. Python 3 is one byte longer for parens in for k in(n,0)].

*Except that 2^a + 2^b with a==b+1 has one longer bit-length, but that's fine because we can interpret that as 2^(a+1)-2^b.

JavaScript (ES6), 73 bytes

(n,[s,f,z]=/^1+(.*1)?(0*)$/.exec(n.toString(2)))=>[s.length-!!f,z.length]

For the subtraction case, the first number is the number of digits in the binary representation and the second number is the number of trailing zeroes. For the addition case, we subtract 1 from the first number. If the binary representation is all 1s followed by some 0s then the addition case is assumed otherwise the subtraction case is assumed. 36-byte port of @xnor's version that only works for B≤30 in JavaScript:

n=>[(l=Math.log2)(n+(n&=-n))|0,l(n)]

Perl, 52 49 32 bytes

Old solution (49 bytes)

Includes +1 for -p

Give input on STDIN:

pow2.pl <<< 17179867136

pow2.pl

#!/usr/bin/perl -p
$_=reverse sprintf"%b",$_;/()1(?:1+|0*)/;$_="@+"

However, using xnor's algorithm and adding a twist gives 32 bytes:

perl -nE 'say 13/9*log|0for$;=$_&-$_,$_+$'

Just the code:

say 13/9*log|0for$;=$_&-$_,$_+$

This suffers from severe rounding error because 13/9 = 1.444... is quite a bit above 1/log 2 = 1.44269... (log itself also has a rounding error but that is so much smaller that we can wrap it up in the analysis of 13/9). But since any 2**big - 2** small gets corrected to 2** big before the log this doesn't mater and the calculation for 2**big + 2 * 2**small gets truncated down so is also safe.. And at the other side of the range 2**n+2**(n-1) doesn't get increased enough in the range [0,64] (I can't properly support more than the integer range anyways due to the use of &) to lead to a wrong result (multiplicator 1.5 however would be too far off for large numbers).