To find all positive integers $a, b$ for which $a^4 + 4b^4$ is a prime number, we first analyze the expression:

$$a^4 + 4b^4$$

This can be rewritten using the Sophie Germain identity:

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

For the expression $a^4 + 4b^4$ to be a prime number, it must be the product of two factors, one of which must be 1, since a prime number only has itself and 1 as positive divisors. Hence, we examine the two cases:

1. $a^2 + 2b^2 - 2ab = 1$ and $a^2 + 2b^2 + 2ab = \text{prime}$

2. $a^2 + 2b^2 + 2ab = 1$ and $a^2 + 2b^2 - 2ab = \text{prime}$

Case 1: If $a^2 + 2b^2 - 2ab = 1$, then

$$a^2 - 2ab + 2b^2 = 1$$

Completing square in $a$, we have

$$(a-b)^2 + b^2 = 1$$

This simplifies to:

$$(a-b)^2 + b^2 = 1$$

For positive integers $a$ and $b$, the viable solution is $(a-b)^2 = 0$ and $b^2 = 1$ which gives $a = b = 1$.

Substituting $a = 1$ and $b = 1$ into the original expression:

$$a^4 + 4b^4 = 1^4 + 4 \times 1^4 = 1 + 4 = 5$$

5 is a prime number.

Case 2: If $a^2 + 2b^2 + 2ab = 1$, the minimum value for both $a^2$ and $b^2$ being positive integers starts from 1, hence making this impossible since the minimum would be more than 1.

Thus, the only possible solution is $(a, b) = (1, 1)$ where the expression results in a prime number.

Therefore, the solution in positive integers for which $a^4 + 4b^4$ is a prime number is:

$$\boxed{(1, 1)}$$