$\mathbf{N 3 3}$ (38-5, Czech) Find all pairs of integers $(a, b)$, where $a \geqslant 1, b \geqslant 1$, that satisfy the equation $a^{b^{2}}=b^{a}$.
Solution: It is obvious that when one of $a, b$ equals 1, $(a, b)=(1,1)$. Now assume $a, b \geqslant 2$.
Let $t=\frac{b^{2}}{a}>0$, then substituting into the given equation we have
$$
a^{\omega}=b^{a},
$$

so $b=a^{t}$, thus $a t=b^{2}=a^{2 t}$, hence $t=a^{2 t-1}$.
If $2 t-1 \geqslant 1$, then
$$
\begin{aligned}
t & =a^{2 t-1} \geqslant(1+1)^{2 t-1} \geqslant 1+(2 t-1) \\
& =2 t>t,
\end{aligned}
$$

which is a contradiction. Therefore, $2 t-1<1$ and $t$ is a rational number. Let $a=b^{k}$, substituting into the given equation we get $b^{k \omega^{2}}=b b^{k^{k}}$, thus
$$
k=b^{k-2} .
$$

If $k \leqslant 2$, then $k=b^{k-2} \leqslant 1$, which is a contradiction. Therefore, $k>2$. Let $k=\frac{p}{q}, p, q \in \mathbf{N}^{*},(p, q)=1$, then $p>2 q$. Thus from (1) we have
$$
\left(\frac{p}{q}\right)^{q}=k^{q}=b^{p-2 q}
$$

is an integer, hence $q^{q} \mid p^{q}$, and since $(p, q)=1$, it follows that $q=1$. Therefore, $k$ is an integer greater than 2.
When $b=2$, from (1) we get $k=2^{k-2}$, so $k \geqslant 4$. Also,
$$
\begin{aligned}
k & =2^{k-2}=(1+1)^{k-2} \\
& \geqslant \mathrm{C}_{k-2}^{0}+\mathrm{C}_{k-2}^{1}+\mathrm{C}_{k-2}^{2} \\
& =1+(k-2)+\frac{(k-2)(k-3)}{2} \\
& =1+\frac{(k-1)(k-2)}{2} \\
& \geqslant 1+(k-1) \\
& =k,
\end{aligned}
$$

where equality holds if and only if $k=4$. Therefore,
$$
a=b^{k}=2^{4}=16 .
$$

When $b \geqslant 3$, we have
$$
\begin{aligned}
k & =b^{k-2} \geqslant(1+2)^{k-2} \geqslant 1+2(k-2) \\
& =2 k-3,
\end{aligned}
$$

thus $k \leqslant 3$, so $k=3$, hence $b=3, a=b^{k}=27$.
In summary, all positive integer pairs $(a, b)$ that satisfy the given equation are $(a, b)=(1,1),(16, 2), (27,3)$.