Solution:
It is clear that $p \neq q$. We set
$$1+\frac{p^{q}-q^{p}}{p+q}=r$$
and we have that
$$p^{q}-q^{p}=(r-1)(p+q)$$
From Fermat's Little Theorem we have
$$p^{q}-q^{p} \equiv -q \quad(\bmod p)$$
Since we also have that
$$(r-1)(p+q) \equiv -r q - q \quad(\bmod p)$$
from (3) we get that
$$r q \equiv 0 \quad(\bmod p) \Rightarrow p \mid q r$$
hence $p \mid r$, which means that $p=r$. Therefore, (3) takes the form
$$p^{q}-q^{p}=(p-1)(p+q)$$
We will prove that $p=2$. Indeed, if $p$ is odd, then from Fermat's Little Theorem we have
$$p^{q}-q^{p} \equiv p \quad(\bmod q)$$
and since
$$(p-1)(p+q) \equiv p(p-1) \quad(\bmod q)$$
we have
$$p(p-2) \equiv 0 \quad(\bmod q) \Rightarrow q|p(p-2) \Rightarrow q| p-2 \Rightarrow q \leq p-2<p$$
Now, from (4) we have
$$p^{q}-q^{p} \equiv 0 \quad(\bmod p-1) \Rightarrow 1-q^{p} \equiv 0 \quad(\bmod p-1) \Rightarrow q^{p} \equiv 1 \quad(\bmod p-1)$$
Clearly $\operatorname{gcd}(q, p-1)=1$ and if we set $k=\operatorname{ord}_{p-1}(q)$, it is well-known that $k \mid p$ and $k<p$, therefore $k=1$. It follows that
$$q \equiv 1 \quad(\bmod p-1) \Rightarrow p-1 \mid q-1 \Rightarrow p-1 \leq q-1 \Rightarrow p \leq q$$
a contradiction.
Therefore, $p=2$ and (4) transforms to
$$2^{q}=q^{2}+q+2$$
We can easily check by induction that for every positive integer $n \geq 6$ we have $2^{n}>n^{2}+n+2$. This means that $q \leq 5$ and the only solution is for $q=5$. Hence the only pair which satisfy the condition is $(p, q)=(2,5)$.