How to show that the following function isn't a polynomial over Q?

For each positive integer $n$ and any rational $x$, we have
$$f(x)\geq (x-b_1)^2(x-b_2)^2\dots(x-b_n)^2.$$ For large $x$, we then have $f(x)\gg x^{2n}$, which implies that if $f$ is a polynomial, it must have degree $≥2n$.

You can adapt the same approach as follows: Say $p_n(x)$ is an $n$-th degree polynomial matching $f(x)$ at all rational points, then in particular, $$ p_n(b_1) = 0\\ p_n(b_2) = (b_2-b_1)^2 \\ \vdots\\ p_n(b_k) = \sum_{i\in\Bbb Z+, i<k}\prod_{j\in\Bbb Z+, j\leq i} (b_k-b_i)^2 $$ For a given fixed sequencing of the rationals as $b_1, b_2, \cdots$, and for any given $k$, the latter expression is just some fixed rational number.

So $p(n)$ is fixed by its values at $b_1 \ldots b_n$, and now consider $-f(b_{n+1}-p_n(b_{n+1}))$. Since all the terms past the $n+2$ term in $f(b_{n+1})$ are zero, $$f(b_{n+1}) = p_n(b_{n+1}) + \prod_{i\leq n}(b_{n+1}-b_i)^2 > p_n(b_{n+1})$$ which contradicts the statement that $f$ matches $p_n$ at all rationals.