If $\gcd(a,b)=1$ , and $a$ is even and $b$ is odd then $\gcd(2^{a}+1,2^{b}+1)=1$? How to prove that:

$\gcd(a,b)=1 \Rightarrow \gcd(2^{a}+1,2^{b}+1)=1$ ,where $a$ is even and $b$ is odd natural number

For example:
$\gcd(2^8+1,2^{13}+1)=1 , \gcd(2^{64}+1,2^{73}+1)=1$
I know that Knuth showed that:
$\gcd(2^{a}-1,2^{b}-1)=2^{\gcd(a,b)}-1$
so: $\gcd(a,b)=1\Rightarrow \gcd(2^{a}-1,2^{b}-1)=1$
but I don't see whether this fact is useful. 

We post a second answer that, instead of doing details, recycles old results. The OP mentioned the following standard result, but was not sure it was useful:
$$\gcd(2^{x}-1,2^{y}-1)=2^{\gcd(x,y)}-1.$$
It is useful! By the result, if $\gcd(a,b)=1$, then 
$$\gcd(2^{2a}-1,2^{2b}-1)=2^{2}-1=3.$$
Thus
$$\gcd((2^{a}+1)(2^a-1)\;,\;(2^b+1)(2^b-1)=3.$$
It follows that any common divisor of $2^a+1$ and $2^b+1$ must divide $3$, and therefore $\gcd(2^a+1,2^b+1)$ is  $1$ or $3$.  
Like in the other answer, we rule out $3$ by using the fact that one of $a$ and $b$, say $a$, is even. For even $a$, $2^a\equiv 1\pmod{3}$, so $3$ does not divide $2^a+1$.
