To find all pairs of positive integers $(x, p)$, where $p$ is a prime, $x \leq 2p$, and $x^{p-1}$ divides $(p-1)^{x} + 1$, we proceed as follows:

1. **Case 1: $x = 1$.**

For $x = 1$, we have $(p-1)^{x} + 1 = (p-1) + 1 = p$. Since $1^{p-1} = 1$, $1$ divides $p$ for any prime $p$. Therefore, $(x, p) = (1, p)$ is a solution for any prime $p$.

2. **Case 2: $x = 2$.**

For $x = 2$, we need to ensure $2^{p-1}$ divides $(p-1)^{2} + 1$. We compute:
$$(p-1)^{2} + 1 = (p-1)^2 + 1 = p^2 - 2p + 1 + 1 = p^2 - 2p + 2.$$
Hence, we need $2^{p-1}$ to divide $p^2 - 2p + 2$.

For $p = 2$, we have $x \leq 2 \times 2 = 4$. Evaluating:
$$2^{2-1} = 2$$
and
$$(2-1)^2 + 1 = 1^2 + 1 = 2.$$
Thus, $2$ divides $2$. Therefore, $(x, p) = (2, 2)$ is a valid solution.

3. **Case 3: $x = 3$.**

For $x = 3$, we need $3^{p-1}$ to divide $(p-1)^{3} + 1$. For $p = 3$, we have $x \leq 2 \times 3 = 6$. Evaluating:
$$3^{3-1} = 9$$
and
$$(3-1)^3 + 1 = 2^3 + 1 = 8 + 1 = 9.$$
Here, $9$ divides $9$. Therefore, $(x, p) = (3, 3)$ is another solution.

After checking these cases, we conclude that the set of solution pairs $(x, p)$ includes:
$$\boxed{(2, 2), (3, 3), (1, p) \text{ for any prime } p}.$$