Suppose $\varphi$ is a sequence $\varphi = \{\varphi(x)\}_{x\in \mathbb{Z}^{d}}$ satisfying the following condition. There exists $k \in \mathbb{N}$ and $C \ge 0$ such that: $$|\varphi(x)| \le C ||x||^{k} \tag{1}\label{1}$$ for every $x \in \mathbb{Z}^{d}$. Here, $||x|| = \sqrt{x_{1}^{2}+\cdots x_{d}^{2}}$.
Since $\mathbb{Z}^{d}$ is countable, I can actuallt treat $\varphi$ to be a sequence indexed by $\mathbb{N}$ instead of $\mathbb{Z}$, say, $\varphi = \{\varphi_{n}\}_{n\in \mathbb{N}}$. This is what I'm trying to prove.
Claim 1: There exists $m \in \mathbb{Z}$ such that: $$\sum_{n\in \mathbb{N}}n^{2m}|\varphi_{n}|^{2} < \infty \tag{2}\label{2}.$$
Claim 2: If (\ref{2}) holds, then (\ref{1}) also hold.
Of course, I have to use condition (\ref{1}) to get (\ref{2}) and vice-versa, but I don't know how exactly does the $||x||$ is afected by the bijection $T: \mathbb{N} \to \mathbb{Z}^{d}$. In other words, when $\{\varphi(x)\}_{x\in \mathbb{Z}^{d}}$ becomes $\{\varphi_{n}\}_{n\in \mathbb{N}}$, how is condition (\ref{1}) changed, so I can use it to prove (\ref{2})?
As an assumption, we will need that $T$ and its inverse grow at most polynomial, that is, there exist a constants $c,\hat C>0$ with $$ \| T(n) \| \leq n^c, \forall n\in\Bbb N \qquad T^{-1}(x) \leq 1+\hat C\|x\|^c, \forall x\in \Bbb Z^d. $$ Such a $T$ exists, see below.
(1) $\implies$ (2): We have $$ \sum_{n\in\Bbb N} n^{2m} |\varphi_n|^2 = \sum_{n\in\Bbb N} n^{2m} |\varphi(T(n))|^2 \leq \sum_{n\in\Bbb N} n^{2m} C^2\|(T(n))\|^{2k} \leq \sum_{n\in\Bbb N} C^2 n^{2m} n^{2kc} =C^2\sum_{n\in\Bbb N} n^{2m+2kc}. $$ If we choose $m$ such that $2m+2kc<-1$, this sum will be finite.
(2) $\implies$ (1): This is not true: If we choose $\varphi$ such that $\varphi(0)=1$ and $\varphi(x)=0$ for all $x\neq0$, then (1) is never satisfied, but (2) is satisfied.
However, if we ignore the $x=0$ case, then this direction can be shown: From (2) we obtain $n^{2m}|\varphi_n|^2\leq1$ for large $n$, or $|\varphi_n|\leq n^{-m}$. Then, one has $$ |\varphi(x)| = |\varphi_{T^{-1}(x)}| \leq T^{-1}(x)^{-m} \leq (1+\hat C\|x\|^c)^{-m} \\\qquad \leq (1+\hat C\|x\|^c)^{\max(-m,0)} \leq \tilde C\| x\|^{\max(-cm,0)} \leq \tilde C\| x\|^{k}, $$ for all but finitely many $x\in\Bbb Z^d$, where $\tilde C$ is a suitable constant and $k>\max(-cm,0)$.
Existence of $T$ with growth conditions: There are many ways to construct such a bijection. One such possibility was already suggested in the comments. A similar possibility is to sort all points $x\in\Bbb Z^d$ according to $\|x\|$, which gives a sequence $x_n$ such that $\|x_n\|$ is non-decreasing. Then we set $T(n)=x_n$. For the first estimate we then have $\|T(n)\| \leq \|(n,0,\dots,0)\| = n \leq n^d$. For the second estimate, let $x\in\Bbb Z^d$ be given. There exists less than $1+C \|x\|^d$ points $y\in\Bbb Z^d$ with $\|y\|\leq \|x\|$, where $C>0$ is a suitable constant. Due to the sorting by the norm, this implies $T^{-1}(x)\leq 1+C\|x\|^d$. In summary, the growth conditions are satisfied with $c=d$.