We say that a sequence $\{b_n\}$ of positive real numbers is good if there exists a positive constant $N$ such that
$$b_1 + b_2 + \cdots + b_n < N b_{n+1} \quad (1)$$
for any $n \in \mathbb{Z}^+$.
We first prove the following equivalent conditions for a sequence to be good.

Claim. A sequence $\{b_n\}$ of positive real numbers is good if and only if the following conditions hold.
(i) There exists a constant $r > 0$ such that $b_{n+1} > r b_n$ for any $n \in \mathbb{Z}^+$; and
(ii) there exists a positive integer $d$ such that $b_{n+d} > 2 b_n$ for any $n \in \mathbb{Z}^+$.

Proof. Suppose $\{b_n\}$ is good. Then we have $N b_{n+1} > b_n$. So we can take $r = \frac{1}{N}$ for (i). Next, we take $d$ to be any integer larger than $2N^2 + 1$. Note that $N b_{n+j} > b_1 + b_2 + \cdots + b_{n+j-1} > b_n$ for any $j \in \mathbb{Z}^+$. It follows that
$$\begin{aligned} b_{n+d} &> \frac{1}{N}(b_1 + b_2 + \cdots + b_{n+d-1}) \\ &> \frac{1}{N}(b_{n+1} + b_{n+2} + \cdots + b_{n+d-1}) \\ &> \frac{1}{N^2}(b_n + b_{n+1} + \cdots + b_n) \\ &= \frac{d-1}{N^2} b_n \\ &> 2 b_n. \end{aligned}$$
This proves (ii).

Conversely, suppose (i) and (ii) hold. For any $n \in \mathbb{Z}^+$, we have
$$\begin{aligned} b_1 + b_2 + \cdots + b_n &= \sum_{j=1}^{d} (b_{n+1-j} + b_{n+1-j-d} + b_{n+1-j-2d} + \cdots) \\ &< \sum_{j=1}^{d} \left( b_{n+1-j} + \frac{1}{2} b_{n+1-j} + \frac{1}{2^2} b_{n+1-j} + \cdots \right) \\ &< \sum_{j=1}^{d} 2 b_{n+1-j} \\ &< 2 \left( \sum_{j=1}^{d} \frac{1}{r^j} \right) b_{n+1}. \end{aligned}$$
So we can take $N = 2 \left( \sum_{j=1}^{d} \frac{1}{r^j} \right)$ so that (1) holds. $\square$

We go back to the original problem. Note that the sequence $\{a_n^2\}$ is good by the given condition. By the claim, there exists $r > 0$ and $d \in \mathbb{Z}^+$ such that $a_{n+1}^2 > r a_n^2$ and $a_{n+d}^2 > 2 a_n^2$ for any $n \in \mathbb{Z}^+$. This implies $a_{n+1} > \sqrt{r} a_n$ and
$$a_{n+2d} > \sqrt{2} a_{n+d} > 2 a_n.$$
Therefore, the sequence $\{a_n\}$ is good by the claim. This is exactly our goal.