To prove the existence of infinitely many integer pairs $(x, y)$ such that $x^{13} + y^{13} \in A$ and $x + y \notin A$, we will proceed as follows:

1. **Understanding the Set $A$**:
The set $A$ is defined as $A = \{a^2 + 13b^2 \mid a, b \in \mathbb{Z}, b \neq 0\}$. This means that any element in $A$ can be written as $a^2 + 13b^2$ for some integers $a$ and $b$ with $b \neq 0$.

2. Quadratic Residue Condition:
For $x \in A$, $-13$ must be a quadratic residue modulo $x$. This can be expressed using the Legendre symbol $\left(\frac{-13}{x}\right) = 1$. Using quadratic reciprocity, we know that $\left(\frac{13}{p}\right) = 1$ if and only if $p \equiv 1, 3, 4, 9, 10, 12 \pmod{13}$ or $p = 2$.

3. **Legendre Symbol for $-13$**:
Similarly, $\left(\frac{-13}{p}\right) = 1$ if and only if $p \equiv 1, 7, 9, 11, 15, 17, 19, 29, 31, 35, 47, 49 \pmod{52}$ or $p = 2$.

4. **Finding $x$ and $y$**:
We need to find $x$ and $y$ such that $x + y$ is not in the set of numbers where $-13$ is a quadratic residue modulo $52$, but $x^{13} + y^{13}$ is in $A$.

5. **Choosing $x$ and $y$**:
Let $x$ be odd and $y$ be even. This ensures that $x + y$ is odd. We need to ensure that $x + y$ is not one of the numbers where $-13$ is a quadratic residue modulo $52$.

6. **Checking $x^{13} \mod 52$**:
For an odd $x$, $x^{13} \equiv x \pmod{52}$. For an even $y$, if $4 \nmid y$, then $y^{13} \not\equiv y \pmod{52}$. For example, $2^{13} \equiv 28 \pmod{52}$.

7. Example:
Let $x \equiv 3 \pmod{52}$ and $y \equiv 2 \pmod{52}$. Then:
$$x^{13} + y^{13} \equiv 3^{13} + 2^{13} \equiv 3 + 28 \equiv 31 \pmod{52}$$
Since $31$ is in the set of numbers where $-13$ is a quadratic residue modulo $52$, $x^{13} + y^{13} \in A$.

8. Ensuring Infinitely Many Pairs:
We can generalize this by considering $x = 3 + 52m$ and $y = 2 + 52n$ for integers $m$ and $n$. This ensures that there are infinitely many such pairs $(x, y)$.

9. Conclusion:
Therefore, there exist infinitely many integer pairs $(x, y)$ such that $x^{13} + y^{13} \in A$ and $x + y \notin A$.

$\blacksquare$