Let the roots of $x^2 + px + q = 0$ be $r$ and $s$, with $r \neq s$ and both integers.

By Vieta's formulas:
$r + s = -p$
$rs = q$

Since $p$ and $q$ are primes, $-p$ is the sum of two distinct integers, and $q$ is their product.

Let $r$ and $s$ be distinct integers. Since $q$ is prime, $rs = q$ implies that one of $r$ or $s$ is $1$ and the other is $q$, or one is $-1$ and the other is $-q$.

Case 1: $r = 1$, $s = q$ (with $q$ prime, $q \neq 1$)
Then $r + s = 1 + q = -p \implies p = -(1 + q)$
But $p$ must be prime, so $-(1 + q)$ is prime. Since $q$ is prime $\geq 2$, $1 + q \geq 3$, so $-(1 + q)$ is negative and prime. The only negative primes are $-2, -3, -5, \ldots$
So $1 + q = 2 \implies q = 1$ (not prime), $1 + q = 3 \implies q = 2$, $1 + q = 5 \implies q = 4$ (not prime), $1 + q = 7 \implies q = 6$ (not prime), etc.
So only $q = 2$ gives $p = -3$ (which is prime).

Check: $x^2 + (-3)x + 2 = x^2 - 3x + 2 = (x - 1)(x - 2)$, roots $1$ and $2$ (distinct integers).

Case 2: $r = -1$, $s = -q$ (with $q$ prime, $q \neq 1$)
Then $r + s = -1 - q = -p \implies p = 1 + q$
$p$ must be prime, so $1 + q$ is prime. Try $q = 2$, $p = 3$ (prime), $q = 3$, $p = 4$ (not prime), $q = 5$, $p = 6$ (not prime), $q = 7$, $p = 8$ (not prime), etc.
So only $q = 2$ gives $p = 3$ (prime).

Check: $x^2 + 3x + 2 = (x + 1)(x + 2)$, roots $-1$ and $-2$ (distinct integers).

Case 3: $r = q$, $s = 1$ (already considered in Case 1).
Case 4: $r = -q$, $s = -1$ (already considered in Case 2).

Therefore, the only pairs $(p, q)$ of prime integers such that the solutions of $x^2 + px + q = 0$ are two distinct integers are:
$$(p, q) = (3, 2) \text{ and } (p, q) = (-3, 2)$$