Solution:

Argue by contradiction. Suppose that for all $i$, $a_{i}=p_{i} / q_{i}$ for some relatively prime positive integers $p_{i}, q_{i}$. Substitute in the given relation:
$$\frac{p_{n+1}^{2}}{q_{n+1}^{2}}=\frac{p_{n}+q_{n}}{q_{n}} \Rightarrow p_{n+1}^{2} q_{n}=q_{n+1}^{2}\left(p_{n}+q_{n}\right)$$
But $p_{n+1}^{2}$ and $q_{n+1}^{2}$ are relatively prime, and $q_{n}$ and $q_{n}+p_{n}$ are also relatively prime (why?). This means that $p_{n+1}^{2}=p_{n}+q_{n}$ and $q_{n}=q_{n+1}^{2}$. In particular, $q_{n+1}=\sqrt{q_{n}}=\sqrt[4]{q_{n-1}}=\cdots=\sqrt[2 n]{q_{1}}$. Unless $q_{1}=1$, we will eventually run out of perfect squares in $q_{1}$ and $q_{n}$ will not be an integer, a contradiction. Thus, the only possibility is $q_{1}=1$, so that all $q_{n}=1$, and our numbers $a_{n}$ are after all integers.

But this is absurd! Indeed, $a_{n+1}=\sqrt{a_{n}+1}<a_{n}$ (because $a_{n}+1<a_{n}^{2}$ for all integers $a_{n}>1$, and $a_{1}=1$ implies an immediate contradiction for $a_{2}=\sqrt{2}$ is irrational), so that we obtain an infinite strictly decreasing sequence of positive integers, and there isn't simply such a thing! This shows that our supposition was wrong and at least one of the $a_{n}$'s will be irrational.