Example 6 Let $n$ be a positive integer of the form $a^{2}+b^{2}$, where $a, b$ are two coprime positive integers. It satisfies: if $p$ is a prime and $p \leqslant \sqrt{n}$, then $p \mid a b$. Find all positive integers $n$ that meet the requirements.
Let $n=a^{2}+b^{2}$ be a number that meets the requirement.
If $a=b$, then by $(a, b)=1$, we know $a=b=1$, in this case $n=2$ meets the requirement.
If $a \neq b$, without loss of generality, assume $a < b$. Take a prime factor $p$ of $b-a$. By $(a, b)=1$, we know $(b-a, b)=(b-a, a)=1$, thus $(b-a, ab)=1$, hence $p \nmid ab$. On the other hand, $p \leqslant b-a < \sqrt{a^{2}+b^{2}} = \sqrt{n}$, which contradicts the requirement that $n$ meets. Therefore, $b-a=1$.

At this point, $n=a^{2}+(a+1)^{2}=2a^{2}+2a+1$. Direct verification shows that when $a=1,2$, $n=5,13$ meet the requirement. When $a \geqslant 3$, if $a$ is odd, then $(a+2, a(a+1))=1$, at this time a prime factor $p$ of $a+2$ satisfies $p \leqslant a+2 \leqslant \sqrt{2a^{2}+2a+1} = \sqrt{n}$, and $p \nmid a(a+1)$, a contradiction; if $a$ is even, then $(a-1, a(a+1))=1$, at this time a prime factor $p$ of $a-1$ satisfies $p \leqslant a-1 < \sqrt{2a^{2}+2a+1} = \sqrt{n}$, and $p \nmid a(a+1)$, a contradiction.

In summary, the only $n$ that meet the condition are 2, 5, and 13.