choose any $(a, b)$ such that $b$ is the smallest. Then the quadratic equation
$$x^2 + (1 - kb)x + b^2 + b = 0$$
has an integral root $x = a$. Let $x = a'$ be the second root, it follows from $a + a' = kb - 1$ that $a' \in \mathbb{Z}$, and from
$$a \cdot a' = b(b + 1)$$
that $a' > 0$. Hence, we have
$$\frac{b+1}{a'} + \frac{a'+1}{b} = k.$$
And it follows from the assumption on $b$ that
$a \geq b, a' \geq b.$
So one of $a$ and $a'$ is equal to $b$. Without loss of generality, we may assume $a = b$, so $k = 2 + \frac{2}{b}$, and so $b \mid 2$ i.e., $b = 1$ or $2$, and $k = 3$ or $4$, respectively.
If $a = b = 1$, then $k = 4$; if $a = b = 2$, then $k = 3$.
Consequently, $k = 3$ or $4$ is the only solution. $\square$