3^2048-1 = (3^1024+1)(3^1024-1) = (3^1024+1)(3^512+1)(3^512-1) =

(3^1024+1)(3^512+1)(3^256+1)(3^128+1)(3^64+1)(3^32+1)(3^16+1)(3^8+1)(3^4+1)(3^2+1)(3+1)(3-1)

3 raised to an even power = 1 mod 4, so each of the first 10 terms is divisible by 2 but not 4.

As for the last two terms,

(3+1) is divisible by 2^2

(3-1) is divisible by 2

So, the highest power of 2 that divides 3^2048-1 is 2^13