Given that $p$ is an odd prime, is the GCD of any two numbers of the form $2^p + 1$ always equal to $3$? I have checked it for some numbers and it appears to be true. Also I am able to reduce it and get the value $3$ for specific primes $p_1$, $p_2$ by using the Euclidean algorithm but I am not able to find a general argument for all numbers.

In general, the following statement is true. Let $a,b$ be positive integers and $m\ge 2$ an integer. Then
$$
\gcd(m^a-1,m^b-1)=m^{\gcd(a,b)}-1.
$$
This can be proved by considering the Euclidean algorithm in base $m$,
see the references here.
Edit: As pointed out by Daniel, one can show in a similar way that
$$
\gcd(m^a+1,m^b+1)\mid m^{2\gcd(a,b)}-1.
$$
For $m=2$ and $p,q$ different odd primes we have equality.
