I will show that the following are equivalent:
Define $v(t) = \sum_k v_k t^k$. I will show the following are equivalent.
(1) One of the roots of $v(t)$ is a root of unity.
(2) There is a nonzero solution $w_k$ where the $w_k$ remain bounded as $k \to \pm \infty$.
(3) There is a nonzero solution $w_k$ where all the $w_k$ are in $\{ -1, 0, 1 \}$.
I will also show that the following are equivalent.
(4) One of the roots of $v(t)$ is a $2^j$-th root of unity, for some $j$.
(5) The polynomial $v(t)$ is divisible either by $1-t$ or by $1+t^{2^{j-1}}$, for some $j$.
(6) The polynomial $v(t)$ is divisible by a polynomial of the form $1-t^m$ or $1+t^m$.
(7) There is a solution $w_k$ where all the $w_k$ are $\pm 1$.
Obviously, the bottom four are most restrictive than the top three.
To knock off the easy ones, clearly $(7) \implies (3) \implies (2)$ and $(4) \implies (1)$. Also, $(4)$ and $(5)$ are easily equivalent: I am just listing the cyclotomic polynomials for $2^j$-th roots of unity. Obviously, $(5) \implies (6)$. The reverse implication $(6) \implies (5)$ is almost as easy: $1-t^n$ is divisible by $1-t$, and, if $n = 2^j (2k+1)$, then $1+t^{2^j}$ divides $1+t^n$.
It is also easy to see that $(6) \implies (7)$: If $1-t^n$ divides $v(t)$ then take any $w_k$ with $w_{k+n} = w_k$; if $1+t^n$ divides $v(t)$ then take any $w_k$ with $w_k=-w_{k+n}$.
The remaining arguments, which will be harder, show $(1) \Longleftrightarrow (2)$, $(1) \implies (3)$ and $(7) \implies (4)$.
We first discuss the proof of $(1) \Longleftrightarrow (2)$, as it sets the context for all the others:
For fixed $v$, the relation $\sum_j v_{j} w_{k+j}=0$ gives a linear recurrence for the $w_i$. The general solutions to this recurrence are of the form
$$w_k = \sum p_i(k) \alpha_i^k \quad (\ast)$$
where the $p_i$ are polynomials and the $\alpha_i$ are the roots of $v(t^{-1})$.
Suppose there is a bounded nonzero expression of the form $(\ast)$. Since we must remain bounded as $k \to \infty$ and $k \to - \infty$, we deduce that all the $\alpha_i$ have norm $1$. But the leading and trailing coefficients of $v(t)$ are $\pm 1$ and all the coefficients of $v(t)$ are integers, so the $\alpha_i$ are roots of unity (read the comments on the linked answer). Conversely, if $\alpha$ is a root of unity and a root of $v(t^{-1})$, then $w_k = \alpha^{-k}$ is a bounded solution to the recurrence. This establishes the equivalence of the first two points.
We now show $(7) \implies (1)$. From our discussion in $(1) \Longleftrightarrow (2)$, any bounded solution must be of the form
$$w_k = \sum c_r \alpha_r^k$$
for $\alpha_r$ various roots of unity that are zeroes of $v(t)$.
We assume that all the $c_r$ are nonzero (otherwise, discard the corresponding $\alpha_r$ from the list).
Let $N$ be the LCM of the orders of the $\alpha_r$, so $w_k$ has period $N$.
Let $N=2^k (2m+1)$.
Suppose, for the sake of contradiction, that none of the $\alpha_r$ have $\alpha_r^{2^k}=1$.
Set $\beta_r=\alpha_r^{2^k}$, so $\beta_r$ is a $(2m+1)$-th root of unity that isn't $1$.
Notice that $\sum_{j=0}^{2m} w_{2^k j}$ is a sum of $2m+1$ numbers which are $\pm 1$, so it is nonzero.
On the other hand,
$$\sum_{j=0}^{2m} w_{2^k j} = \sum_{j=0}^{2m} \sum_r c_r \sum_{j=0}^{2m} (\alpha_r^{2^k})^j = \sum_{j=0}^{2m} \sum_r c_r \beta_r^j = \sum_{j=0}^{2m} \sum_r c_r \frac{\beta_r^{2m+1}-1}{\beta_r-1} = 0$$
where the last equality is because $\beta_r$ is a $(2m+1)$-th root of unity that isn't $1$.
The last, and hardest, argument is that $(1) \implies (3)$. Not strictly needed to answer your question, but I am proud of it.
Let $\alpha$, a primitive $n$-th root of unity, be a root of $v(t^{-1})$. Since the coefficients of $v(t)$ are integers, and the $n$-th cyclotomic polynomial is irreducible, all of the primitive $n$-th roots of unity are roots of $v(t)$. We will find a sequence $w_k$ so that all $w_k \in \{ -1, 0, 1 \}$ and $w_k$ can be written as
$$w_k = \sum_{r,\ GCD(r,n)=1} c_r \alpha^{rk} \quad (\dagger)$$
for some constants $c_r$.
Define the polynomial
$$f(t) = \prod_{p | n} (1-t^{n/p})$$
where the product is over primes dividing $n$.
Lemma Every coefficient of $f$ is $\pm 1$, and all of the exponents of $f$ are distinct modulo $n$.
Proof Let $r$ be the number of distinct primes dividing $n$. We claim that all $2^n$ exponents occurring in the product are distinct modulo $n$. Suppose to the contrary that $P$ and $Q$ are two different sets of primes dividing $n$ such that
$$\sum_{p \in P} \frac{n}{p} \equiv \sum_{p \in Q} \frac{n}{p} \bmod n. \quad (\mathbb{S})$$
Since $P \neq Q$, we may assume that there is some $p$ in $P$ and not in $Q$ (or vice versa.) Let $p^e$ be the precise power of $p$ dividing $n$. Then reducing $(\mathbb{S})$ modulo $p^e$ gives
$$p^{e-1} + 0 + \cdots + 0 \equiv 0+ \cdots + 0 \bmod p^{e-1}$$
an absurdity. $\square$
Let $g(t)=\sum_{k=0}^{n-1} g_k t^k$ be the polynomial formed from $f(t)$ by reducing the exponents of $f(t)$ modulo $n$ to lie between $0$ and $n-1$, while keeping the coefficients the same. Let $w_k$ be the sequence formed by repeating the coefficients of $g(t)$ periodically modulo $n$. So $w_k \in \{ -1, 0, 1 \}$, and we must check that $w_k$ can be written in the form $(\dagger)$.
So we have
$$g_k = \sum_r c_r \alpha^{rk}$$
where $c_r$ is given by the discrete Fourier transform
$$c_r = \frac{1}{n} \sum g_k \alpha^{-rk}.$$
If $r$ is NOT relatively prime to $n$, then $\sum g_k \alpha^{-rk} = g(\alpha^{-r}) = f(\alpha^{-r})=0$. So the only nonzero Fourier coefficients are the ones with $GCD(r,n)=1$, and $w_k$ is of the form $(\dagger)$, as desired.
