Solution:

Let us consider $b$; since $a+1$ divides $b-1$ we can write $b = k(a+1) + 1$, with $k$ a natural number. Let us reformulate the second condition as $(k(a+1)+1) h = a^{2} + a + 2$, that is $k h(a+1) + h = a(a+1) + 2$ for some $h \in \mathbb{N}$.

Let us first suppose that $k$ is positive; if $k h = a$, then $h = 2$; from this follow the solutions of the type $(2k, k(2k+1)+1)$ where $k$ is a positive integer. If instead $k h < a$, then $h > a+3$; on the other hand, if $k$ is positive, we must have $h \leq h k$, which is absurd.

There remains the case where $k$ is $0$, and hence $b = 1$; all pairs $(a, 1)$, with $a$ a positive integer, are indeed solutions.

Solution:

As before, let us write $b = k(a+1) + 1$, with $k \in \mathbb{N}$ and $(k(a+1)+1) h = a^{2} + a + 2$ for some $h \in \mathbb{N}$.

We observe that for $b = 1$ we have $k = 0$ and $a$ can be any natural number $n > 0$, that is, all pairs $(n, 1)$ are solutions.

Let us consider the second condition modulo $(a+1)$: we have that $h \equiv 2 \bmod (a+1)$. Now, if we had $h \geq (a+1)+2$, excluding the case $k=0$ already considered, we would get
$$(k(a+1)+1) h \geq (a+2) h \geq (a+2)(a+3) = a^{2} + a + 2 + 4(a+1) > a^{2} + a + 2$$
in contrast with the second condition. Hence $h = 2$ and $2k(a+1) + 2 = a(a+1) + 2$, from which $a = 2k$. Consequently, substituting into the first condition, we have $b = k(2k+1) + 1$, from which we obtain the other solutions $\left(2k, 2k^{2} + k + 1\right)$, for $k > 0$ a natural number.