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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 $$\begin{gathered} ab\ge n^2(n^2+1)\ge n^4+1 \\ (b-a{)}^4+1\le (2n{)}^4+1=16n^4+1 \end{gathered}$$ 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 $1(\mathrm{mod}8)$ 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.