Finding the prime factorization of $2^{22}+1$? The problem is

Find the prime factorization of $2^{22}+1$

I have a solution but I think there must be some better ways:
My Solution:
$2^{22}+1 = (2^{22} + 2 \cdot 2^{11} +1) - 2 \cdot 2^{11} = (2^{11}+1)^2 - 2^{12}$
and we can factor as
$(2^{11}+1-2^6)(2^11+1+2^6) = (2048+1-64)(2048+1+64)= 1985 \cdot 2113$
From here it's not infeasible by hand, but I think there must be a better method.

Suppose $p$ is a prime factor of $2^{22}+1$ greater than $5$.  Then $2^{22}\equiv -1 \pmod{p}$.  So we know the order of $2$ modulo $p$ is $44$, and thus $44 \mid p-1$.  We conclude that $p = 44n+1$ for some $n$.  
Using your factorization, we have that $\sqrt{2113} = 45.96\ldots$.  There are no primes of the form $44n+1$ in the range between $5$ and $45$, so $2113$ is prime.  To factor $1985$, the $5$ is easy, and we're left with $397$.  Again, there are no primes of the form $44n+1$ in the range $5$ to $\sqrt{397}$, so it's prime too.
