Showing that $2^6$ divides $3^{2264}-3^{104}$ 
Show that $3^{2264}-3^{104}$ is divisible by $2^6$. 

My attempt: Let $n=2263$. Since $a^{\phi(n)}\equiv 1 \pmod n$ and
$$\phi(n)=(31-1)(73-1)=2264 -104$$
we conclude that $3^{2264}-3^{104}$ is divisible by $2263$. 
I have no idea how to show divisibility by $2^6$.

A variant with some details:
By *Euler's theorem, $\;3^{\varphi(2^6)}\equiv 3^{32}\equiv=1\mod 2^6$. So
$$3^{2264}-3{104}\equiv3^{2264\bmod\varphi(2^6)}-3^{104\bmod\varphi(2^6)}\equiv 3^{24}-3^{8}\equiv3^8(3^{16}-1)\mod 2^6,$$
so it is enough to prove $3$ has order $16$ mod $2^6=64$. 
Let's use the fast exponentiation algorithm mod $64$:
$$\begin{array}{r| l}
k & 3^k \\\hline
1&3\\2&9\\4&81\equiv 17\\8&289 \equiv-31\\16&961\equiv\color{red}{1}
\end{array}$$
A nicer and faster computation (after @Joffan's suggestion):
As $x\equiv a\mod 2^k\implies x^2\equiv a^2\mod 2^{k+1}$, we have
$$ 3 \equiv -1\mod 2^2\implies3^2 \equiv 1\mod2^3,$$
whence $3^{16}\equiv 1\mod 2^6$.
