Solve the congruence using Pohlig-Hellman algorithm Use the Pohlig-Hellman algorithm to compute a solution to: 
$3^x\equiv 2 \pmod {65537}$

My attempt:

$p-1 = 65537-1 = 65536= 2^{16}$

$x= 2^0x_0+2^1x_1+2^2x_2+...+2^{15}x_{15}$ 
For $x_0$:
$2^{65536/2}=3^{(65536/2)x_0}$
$1 \pmod {65537} \equiv (-1)^{x_0} \pmod{65537}$
So $x_0 = 0$

Is this right? Because for all other steps after that, I get $0$'s for all $x_i$'s where $i=1,2,3,...,15$

Any help would be appreciated!
You are correct in the procedure.
You need to express $x$ as $x = 2^0x_0+2^1x_1+2^2x_2+...+2^{15}x_{15}$ 
Where the $x_i$ are in $GF(2)$

First compute:
$3^{(0*(p-1)/2)} \pmod p \equiv 1$
$3^{(1*(p-1)/2)} \pmod p \equiv 65536$

So that when you'll compute 
$2^{65536/2^{i+1}} \pmod p$ you'll get either 1 (corresponding to $x_{i} = 0$) or 65536 (corresponding to $x_{i} = 1$)

For the first values of $i$ you will get only $x_i=0$, but then for $i=11$: $2^{65536/2^{12}} \pmod p\equiv 65536$ which corresponds to $x_{11}=1$

Then you need to proceed... but I just wanted to show that you don't get 0 for all $x_i$