$p = 2$ and $p$ divides $q + 6$ together imply $2$ divides $q$. Hence $q = 2$. But then $q$ divides $p + 7$ is not satisfied. Thus $p$ is odd.
$q = 2$ and $p$ divides $q + 6$ together imply $p = 2$ but $p = 2 = q$ contradicts $q$ divides $p + 7$. Thus $p$ and $q$ are both odd.
$p + 7$ is even and so $q \le (p + 7)/2 \le (q + 6 + 7)/2$. Hence $q \le 13$.
Now examine each of the cases $q = 3, 5, 7, 11, 13$. Only one case $q = 13$ provides a solution in which $p = 19$.

Second solution: By hypothesis, there are $a, b \in \mathbb{N}$ such that $q + 6 = pa, p + 7 = bq$. Hence
$$13 = p(a-1) + q(b-1) \equiv pm + qn,$$
where $m, n$ are nonnegative integers. Suppose $m, n \in \mathbb{N}$. Then
$$4 \le p+q \le pm+qn = 13,$$
and so $p, q \in \{2, 3, 5, 7, 11\}$. An inspection of the possible pairs $(p, q)$ that can be formed from this set shows that no such pair satisfies the hypotheses. Hence, one of $m, n$ is zero. Suppose $n = 0$. Then $b = 1, a = 2, p = 13$ which means that $q = 20$, which isn't a prime number. It follows that $m = 0$, i.e., $a = 1, b = 2, q = 13$ and hence $p = 26 - 7 = 19$. Thus $p = 19, q = 13$ is the only solution pair.