Series

$\begin{array}{rcl} 2&=&2\times(1^2-0)\\ 9&=&3\times(2^2-1)\\ 28&=&4\times(3^2-2)\\ 65&=&5\times(4^2-3)\\ 126&=&6\times(5^2-4) \end{array}$

Therefore the next number is $7\times(6^2-5)=7\times31=217$ .

Rather an odd approach. Take the difference between any two consecutive elements, we have new series: 7, 19, 37, 61. Again, repeat the same process for this new found series: 12, 18, 24. And repeat again, we then realize that it's 6, 6, so the next element should be 6. Thus, the next element of "12, 18, 24" series should be 30, and it is 91 for "7, 19, 37, 61", and finally the next element of the main series is 126 + 91 = 217. Guess I got it by luck :)

The second differences form an arithmetic sequence. So these are from a cubic polynomial. Pop them into Wolfram-Alpha. (OK, I'm mentally lazy.) Verify its result with Python, using:

[n**3+1 for n in range(1,7)]