Denote $\displaystyle S = \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_{250}}$, it suffices to show $$10 < S < 100.$$ We compute $a_1 = 2$, $a_2 = \frac{9}{4}$, $a_3 = \frac{793}{324} \in \left(\frac{12}{5}, \frac{5}{2}\right)$ and so $a_4 = a_3 + \frac{1}{a_3^2} > \frac{12}{5} + \frac{4}{25} = \frac{64}{25}$. All terms are positive, so $a_{n+1} > a_n$ for $n\geq 1$; in particular, $a_n \geq a_4 > \frac{64}{25}$ for $n=4,5,\ldots,250$. Therefore $$S < \frac{1}{2} + \frac{4}{9} + \frac{2}{5} + 247 \cdot \frac{25}{64} < 2 + 98 = 100.$$ On the other hand, $a_n > a_2 = \frac{9}{4}$ for $n\geq 2$, we have $a_{n+1} < a_n + \frac{16}{81} < a_n + \frac{1}{5}$ for $n\geq 2$.Hence (by induction) $a_n < \frac{n+10}{5}$ for all $n\geq 2$, so $$S > \frac{1}{2} + 5 \cdot \left(\frac{1}{12} + \frac{1}{13} + \frac{1}{14} + \cdots + \frac{1}{260}\right).$$ The latter expression contains $8$ fractions over $\frac{1}{20}$, $30$ other fractions over $\frac{1}{50}$, $50$ other fractions over $\frac{1}{100}$ and $100$ other fractions over $\frac{1}{200}$. Therefore the expression exceeds $5\left(\frac{8}{20} + \frac{3}{5} + \frac{1}{2} + \frac{1}{2}\right) = 10$. In conclusion, $S > 10$.

Thanks to @dangerousliri and @Marinchoo for discussing the following. Direct verification shows that $x + \frac{1}{x^2}$ increases for $x\geq 2$, hence one may show (by induction) a significantly better bound for $a_n$, say. $\sqrt[3]{3n+5} \leq a_n \leq \sqrt[3]{4n+4}$ for all $n$. In other words, $a_n$ grows at rate of $n^{1/3}$. A more direct way to see this is as follows: $a_{n+1}^3 = a_n^3 + 3 + \frac{3}{a_n^3} + \frac{1}{a_n^8}$, so $a_{n+1}^3 - a_n^3 \in [3,4]$ for all $n$, whence (by induction) $\sqrt[3]{3n+5} \leq a_n \leq \sqrt[3]{4n+4}$. In a similar fashion one obtains that for all $\varepsilon > 0$ and sufficiently large $n$ one has $\sqrt[3]{3n} < a_n < \sqrt[3]{(3+\varepsilon)n}$.