Suppose that $x_i,z_i$ are i.i.d and that the moments $\mathbb{E}x_i^m \leq K \mathbb{E} z_i^m$ uniformly for $m=1,2,\ldots$. For simplicity's sake take $x_i$ and $z_i$ to be symmetric so we can ignore the odd moments.
Is it possible to get bound $$\mathbb{E}(a_1x_1 + \cdots + a_nx_n)^{2m} \leq \mathcal{O}(K)\mathbb{E}(a_1z_1+\cdots+a_nz_n)^{2m}$$ that doesn't depend on the length of the sum $n$?
If $I_{2m} = \biggl\{2i_1,2i_2,\ldots,2i_n : i_j \geq 0,\,\sum_{j=1}^n 2i_j = 2m\biggr\}$ then we can write $$\mathbb{E}(a_1x_1 + \cdots + a_nx_n)^{2m} = \sum_{2i_1,\ldots\in I_{2m}} \prod_{j=1}^n a_j^{2i_j} \mathbb{E}x_j^{2i_j} \leq \sum_{2i_1,\ldots\in I_{2m}} \prod_{j=1}^n a_j^{2i_j} K\mathbb{E}z_j^{2i_j} = K^n\mathbb{E}(a_1z_1+\cdots+a_nz_n)^{2m},$$ for instance.