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The only solutions are $(1, 1)$, $(1, 2)$, and $(2, 1)$, which clearly work. Now we show there are no others.

Obviously, $\gcd(a, b) = 1$, so the problem conditions imply
$$ab \mid (a - b)^4 + 1$$
since each of $a$ and $b$ divide the right-hand side. We define
$$k \stackrel{\text{def}}{=} \frac{(b - a)^4 + 1}{ab}.$$

Claim (Size estimate) — We must have $k \le 16$.

Proof. Let $n = \lfloor\sqrt{a}\rfloor = \lfloor\sqrt{b}\rfloor$, so that $a, b \in [n^2, n^2 + 2n]$. We have that
$$ab \ge n^2(n^2 + 1) \ge n^4 + 1 \\ (b-a)^4 + 1 \le (2n)^4 + 1 = 16n^4 + 1$$
which shows $k \le 16$. $\square$

Claim (Orders argument) — In fact, $k = 1$.

Proof. First of all, note that $k$ cannot be even: if it was, then $a, b$ have opposite parity, but then $4 \mid (b-a)^4 + 1$, contradiction.
Thus $k$ is odd. However, every odd prime divisor of $(b-a)^4 + 1$ is congruent to $1 \pmod{8}$ and is thus at least $17$, so $k = 1$ or $k \ge 17$. It follows that $k = 1$. $\square$

At this point, we have reduced to solving
$$ab = (b - a)^4 + 1$$
and we need to prove the claimed solutions are the only ones. Write $b = a + d$, and assume WLOG that $d \ge 0$: then we have $a(a + d) = d^4 + 1$, or
$$a^2 - da - (d^4 + 1) = 0.$$
The discriminant $d^2 + 4(d^4 + 1) = 4d^4 + d^2 + 4$ must be a perfect square.

* The cases $d = 0$ and $d = 1$ lead to pairs $(1, 1)$ and $(1, 2)$.
* If $d \ge 2$, then we can sandwich
$$(2d^2)^2 < 4d^4 + d^2 + 4 < 4d^4 + 4d^2 + 1 = (2d^2 + 1)^2,$$
so the discriminant is not a square.

The solution is complete.