First, we have
$$(p+1)^{p-1} + (p-1)^{p+1} \ge (p+1)^{p-1} \ge (p-1)^{p-1} \quad (1)$$
$$(p+1)^{p-1} + (p-1)^{p+1} < (p+1)^{p+1} + (p+1)^{p-1} = 2(p+1)^{p+1} < (p+2)^{p+2} \quad (2)$$
From (1) and (2) follows
$$(p-1)^{p-1} \le q^q < (p+2)^{p+2}.$$
1) Let
$$q = p-1, \quad (p+1)^{p-1} + (p-1)^{p+1} = (p-1)^{p-1}.$$
From
$$(p+1)^{p-1} + (p-1)^{p+1} \ge (p+1)^{p-1} \ge (p-1)^{p-1}$$
we have $(p-1)^{p+1}=0$ and $(p+1)^{p-1}=(p-1)^{p-1} \Rightarrow p=1, q=0$. But $0$ is not a natural number, so $(p+1)^{p-1}+(p-1)^{p+1}=(p-1)^{p-1}$ has no solution in the set of natural numbers $\mathbb{N}$.

2) Let $q = p$, $(p+1)^{p-1} + (p-1)^{p+1} = p^p$.

If $p=1$, then
$$(p+1)^{p-1} + (p-1)^{p+1} = 1 \text{ и } p^p = 1.$$
So $(p,q)=(1,1)$ is a solution of $(p+1)^{p-1}+(p-1)^{p+1}=q^q$.
If $p=2$, then
$$(p+1)^{p-1} + (p-1)^{p+1} = 4 \text{ и } p^p = 4.$$
Hence $(p,q)=(2,2)$ is a solution of $(p+1)^{p-1}+(p-1)^{p+1}=q^q$.
If $p=3$, then
$$(p+1)^{p-1} + (p-1)^{p+1} = 32 \text{ а } p^p = 27.$$
Hence $(p,q)=(3,3)$ is not a solution of $(p+1)^{p-1}+(p-1)^{p+1}=q^q$.
If $p \ge 4$, it holds $(p-1)^p > p^{p-1}$. We obtain
$$(p+1)^{p-1} + (p-1)^{p+1} > (p+1)^{p-1} + p^{p-1}(p-1) > p^{p-1} + p^{p-1}(p-1) = p^p.$$
Hence, when $p \ge 4$, $(p+1)^{p-1}+(p-1)^{p+1} = p^p$ doesn't have a solution in $\mathbb{N}$.

3) Let $q = p+1$
$$(p+1)^{p-1} + (p-1)^{p+1} = (p+1)^{p+1} \Leftrightarrow$$
$$(p-1)^{p+1} = (p+1)^{p-1}((p+1)^2 - 1) \Leftrightarrow$$
$$(p-1)^{p+1} = (p+1)^{p-1} p(p+2)$$
Since $p$ and $p-1$ are coprime, the equation $(p+1)^{p-1}+(p-1)^{p+1}=(p+1)^{p+1}$ doesn't have a solution in $\mathbb{N}$.

Finally, the solutions of $(p+1)^{p-1}+(p-1)^{p+1}=q^q$ are $(p,q)=(1,1)$ and $(p,q)=(2,2)$.