And $b^{2}+a \mid a^{2}+b$.

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And $b^{2}+a \mid a^{2}+b$.
Solve $b^{2}+a=a^{2}+b$ which is $(a-b)(a+b-1)=0$, so $b=a$. This contradicts the given condition. Therefore, $b^{2}+ab$.

Let $b^{2}+a=p^{k}$, where $p$ is a prime. $a^{2}+b-a\left(a+b^{2}\right)+b^{2}\left(a+b^{2}\right)=b+b^{4}=b\left(b^{3}+\right.$ 1) is divisible by $p^{k}$. Since $\left(b, b^{3}+1\right)=1$, it follows that $p^{k} \mid b$ or $p^{k} \mid\left(b^{3}+1\right)$. But $b<b^{2}+a=p^{k}$, so it must be that $p^{k} \mid\left(b^{3}+1\right)$.

Since $b^{3}+1=(b+1)\left(b^{2}-b+1\right)$, and both $b+1$ and $b^{2}-b+1$ are less than $b^{2}+a$ which is $p^{k}$, it follows that both $b+1$ and $b^{2}-b+1$ are divisible by $p$. And $\left(b+1, b^{2}-b+1\right)=(b+1,2 b-1)=(b+1$, 3). Therefore, $p=3$. Since $b^{2}-b+1=(b+1)^{2}-3(b+1)+3$ leaves a remainder of 3 when divided by 9, we have $\left\{\begin{array}{l}b+1=3^{k-1} t_{1}, \\ b^{2}-b+1=3 t_{2},\end{array}\right.$ where $t_{2}$. Thus, $3(b+1) \geqslant 3^{k}=b^{2}+a \geqslant b^{2}+b+1$. Therefore, $b=$ $2, k=2, a=5$. The only positive integer pair is $(5,2)$.