Solution 1. If $(a, b)$ is a solution with $a < b$, then so is $(b, c)$ where
$$c = \frac{b^2 + 1}{a}.$$
It is immediate that $c \mid b^2 + 1$ so we have to show only that $b \mid c^2 + 1$. We can calculate:
$$a^2(c^2 + 1) = (b^2 + 1)^2 + a^2 = b^2 + 2b + (a^2 + 1) \equiv 0 \pmod{b}$$
As $b \mid a^2 + 1$ then $b$ and $a^2$ are relatively prime, so $b \mid a^2(c^2 + 1)$ implies that $b \mid c^2 + 1$. To complete the construction we just need to observe that $a < b$ implies $ab < b^2 < b^2+1$ and so $b < c$. We can then use this rule and produce an infinitude of increasing solutions starting with $(a, b) = (1, 2)$.

Solution 2. Let $F_n$ be $n^{th}$ Fibonacci number, so that $F_0 = 0, F_1 = 1, F_2 = 1$ and $F_n = F_{n-1} + F_{n-2}$. Catalan's identity, shown by induction, states that:
$$F_n^2 - F_{n+r}F_{n-r} = (-1)^{n-r}F_r^2.$$
Putting $r = 2$ and $n$ odd gives $F_n^2 + 1 = F_{n+2}F_{n-2}$ which provides us with an infinitude of solutions of the form $(a, b) = (F_n, F_{n+2})$.