GCD

The first number $(63^2-4)71$ can be expressed as $(63+2)(63-2)71=5\times 13\times 61 \times 71$

The second number is $(31^2)^2+4(81)= 31^4 + 4(3^4)$

And by the Sophie Germain Identity , we get that

$a^4+4b^4 = (a^2+2b^2+2ab)(a^2+2b^2-2ab)$

Hence the $2^{nd}$ number is $31^4+4(3^4) = (31^1 + 2\times 3^2 + 2\times 31\times 3)(31^1 + 2\times 3^2 - 2\times 31\times 3)$

$31^4+4(3^4) = (961+18 + 186)(961+18-186)=(1165)(793)$

We can factorize $1165$ as $5\times 233$ and $793$ as $13 \times 61$

Hence the two numbers we have are $5\times 13\times 61 \times 71$ and $5\times 13\times 61 \times 233$

(Observe that $71\nmid 233$ hence we are done)

Thus the GCD of the numbers is $5\times 13\times 61$ and the sum of all prime numbers is $5+13+61 = \boxed{79}$