First few elements of the sequence are
$a_1 = 2$, $a_2 = 2(2+2) = 8$, $a_3 = 2(3+8) = 22$, $a_4 = 2(4+22) = 52$, $\dots$
Let us look at the differences $2^{n+2} - a_n$:
$$\begin{align*} 2^3 - a_1 &= 8 - 2 = 6, \\ 2^4 - a_2 &= 16 - 8 = 8, \\ 2^5 - a_3 &= 32 - 22 = 10, \\ 2^6 - a_4 &= 64 - 52 = 12. \end{align*}$$

We assume that $a_n = 2^{n+2} - 2(n+2)$ and we will prove it using mathematical induction.
The basis was already checked.
Assume that $a_n = 2^{n+2} - 2(n+2)$ for some $n \in \mathbb{N}$.
Using the given recursion we get
$$\begin{align*} a_{n+1} &= 2((n+1) + a_n) = 2(n+1+2^{n+2}-2(n+2)) \\ &= 2 \cdot 2^{n+2} - 2(n+3) = 2^{n+3} - 2(n+3). \end{align*}$$
This completes the inductive step and we conclude that for all $n \in \mathbb{N}$
$$a_n = 2^{n+2} - 2(n+2) < 2^{n+2}.$$