For prime numbers $p$ consider the two conditions
$({\rm A})\ \ \forall\:\! k\geq0\!:\ (-2)^k+1\not\equiv0\,\pmod{\!p}$
$({\rm B})\ \ \exists\:\! k\geq0\!:\ \ 2^{2k+1}+1\equiv0\,\pmod{\!p}$
Experimentation seems to show that the odd primes satisfying $\rm(A)$ are precisely the odd primes satisfying $\rm(B)$. Is this correct?
On
Note that lulu's comment has the right general idea. First, note $p = 2$ satisfies condition A, but not condition B. Thus, considering only odd primes $p$, we use the multiplicative order of $-2$ modulo $p$ (rather than of $2$), i.e.,
$$g = \operatorname{ord}_p(-2) \tag{1}\label{eq1A}$$
If $g$ were even, then with $k = \frac{g}{2}$, we'd have
$$(-2)^{k} \equiv \pm 1 \pmod{p} \tag{2}\label{eq2A}$$
Condition A shows that $(-2)^{k} \not\equiv -1 \pmod{p}$, so $(-2)^{k} \equiv 1 \pmod{p}$, but then $1 \le k \lt g$. However, $g$ would then not be the multiplicative order. This contradiction means $g$ must be odd, i.e., it's of the form
$$g = 2k + 1, \;\; k \ge 0 \tag{3}\label{eq3A}$$
We then have
$$(-2)^{2k+1} \equiv 1 \pmod{p} \;\to\; 2^{2k+1} \equiv -1 \pmod{p} \;\to\; 2^{2k+1} + 1 \equiv 0 \pmod{p} \tag{4}\label{eq4A}$$
i.e., condition B.
For the other direction, from the first modulo equation in \eqref{eq4A}, we have condition B meaning that $g$ in \eqref{eq1A} must be odd. Assume there's an integer $k$ where we have
$$(-2)^k + 1 \equiv 0 \pmod{p} \;\;\to\;\; (-2)^{2k} \equiv 1 \pmod{p} \tag{5}\label{eq5A}$$
Thus, we have $g \mid 2k$. However, since $g$ is odd, this means $g \mid k$, but then
$$(-2)^k \equiv 1 \pmod{p} \;\;\to\;\; (-2)^k - 1 \pmod{p} \tag{6}\label{eq6A}$$
This contradicts \eqref{eq5A} unless $2 \equiv 0 \pmod{p}$, but since we're considering only odd primes $p$, this isn't true. Thus, we get that condition B also implies condition A.
In conclusion, we have condition A and condition B implying each other for odd primes $p$, so your hypothesis is correct.
$\rm\small (A),\:\!(B)\:\!$ are both equivalent to: $\,{-}2\:\!$ has odd order, since their logical negations are equivalent to $\:\!(2),(3)\:\!$ in the below characterizations of elements having even order (OP is case $\,a = -2)$.
Theorem $ $ TFAE modulo $\rm\color{#c00}{prime}$ $\,\color{#0af}{p\neq 2},\,$ and $\,\color{#b70}{p\nmid a}$
$(1)\ \ \ {\bf 2}\mid o(a) :=\,$ order of $\,a\bmod{p},\,$ $ $ i.e. $ $ the order of $\,a\,$ is $ $ even
$(2)\ \ \ \color{#0a0}{a^k\,\equiv -1}\ $ for some $\,k\ge 0,\ \, $ i.e. some power $a^k\:\!$ has order $\bf 2$
$(3)\ \ \ a^{2j+1}\!\not\equiv 1\ \,$ for $\ $ all $\ $ $\,j\ge 0,\ \ $ i.e. $\ \ a^n\,\equiv\, 1\ \Longrightarrow\ n\,$ is even
Proof $\ \ (3\!\Rightarrow\!1)\,\ \ \color{#b70}{p\nmid a}\Rightarrow a^{o(a)}\!\equiv 1\, $ so $(3)\Rightarrow o(a)\neq 2j\!+\!1$ $\Rightarrow o(a)\,$ even.
$(1\!\Rightarrow\! 2)\,\ \ o(a)\!=\!2k\Rightarrow a^{2k}\equiv \color{#90f}{1\not\equiv a^k}\,$ so $\,0\equiv (\color{#90f}{a^k-1})(\color{#0a0}{a^k+1})\,\smash{\color{#c00}{\overset{\rm prime}\Rightarrow}} \color{#0a0}{a^k+1}\equiv 0$.
$(2\!\Rightarrow\!3)\ \, $ If $\,a^{2j+1}\!\equiv 1\,$ then $\,o(a)\,|\, 2j\!+\!1\,$ so odd $\:\!\underbrace{o(a)\,|\,2k}_{\color{#0a0}{\textstyle a^{2k}\equiv 1}}\Rightarrow o(a)\mid k\Rightarrow \underbrace{1\equiv \color{#0a0}{a^k\equiv -1}}_{\textstyle\Rightarrow\color{#0af}{p\mid 2\Rightarrow\!\Leftarrow}}$
Remark $ $ The proof works in any integral $\rm\color{#c00}{domain}$ where $\,\color{#0af}{2\neq 0}\,$ and $\,a\,$ has $\rm\color{#b70}{finite \ order}$. Further we can view the above Theorem more conceptually as the special case $\,d=2\,$ of the following characterizations of divisors $\,d\,$ of the order of $\,a.\,$ As in $\,(1\!\Rightarrow2)\,$ above, in an integral $\rm\color{c00}{domain}$ where $\,2\!\neq\!0\,$ we have $\,\color{#0a0}{o(a^k)=2\!\iff\! a^k = -1},\,$ so $(2)$ below is the same as $(2)$ above. Ditto for $\:\!(3)\!:\,$ the contrapositive of $(3)$ below for $\,d=2\,$ is $\,2\nmid n\:\!\Rightarrow\:\! a^n\neq 1,\,$ which is $(3)$ above. These basic results are clarified when one studies cyclic groups and (principal) ideals and modules.
Theorem $ $ TFAE for an element $\,a\,$ of finite order in $\,\Bbb Z_n\,$ (or any monoid).
$(1)\ \ \ d\mid o(a) := $ order of $\,a$
$(2)\ \ \ \color{#0a0}{o(a^k) =d}\,$ for some $\,k\ge 0$
$(3)\ \ \ a^{n}= 1\,\Rightarrow\, d\mid n$
Proof $\ \ (1\!\Rightarrow\! 2)\,\ \ o(a)\!=\!dk\,\Rightarrow\, o(a^k)=d\,$ by here.
$(2\!\Rightarrow\!1)\ \ $ by $\,o(a^k)\mid o(a),\,$ i.e. $\,{a^{o(a)}\equiv 1}\Rightarrow (a^k)^{o(a)}\equiv 1\Rightarrow \color{#0a0}{d\ \mid\ }o(a)$
$(1\!\Rightarrow\! 3)\ \ \ a^n = 1\Rightarrow d\mid o(a)\mid n $
$(3\!\Rightarrow\! 1)\ \ \ a^{o(a)} = 1\Rightarrow d\mid o(a)$
Below is an additive analog for fractions. Call $\,d\neq 0 \,$ a denominator of a fraction $\,q\,$ if $\,dq \in \Bbb Z,\,$ i.e $\,q = c/d\,$ for some $\,c\in \Bbb Z$. The least denom $\,d\,$ is the order of $\,q\,$ in the unit fraction circle group $\,\Bbb Q/\Bbb Z = $ fractions $\!\bmod 1,\,$ i.e. the least $\,d>0\,$ with $\,dq\equiv 0\pmod{\!\Bbb Z}$.
Theorem $ $ TFAE for a fraction $\,a\,$ over $\,\Bbb Z\,$ (or any PID) $\, $ [compare to prior Theorem]
$(1)\ \ \ d\mid \delta(a) := $ least denominator of $\,a$
$(2)\ \ \ \color{#0a0}{\delta(ka) =d}\,$ for some $\,k\ge 0$
$(3)\ \ \ na\in\Bbb Z\,\Rightarrow\, d\mid n,\,$ i.e. $\,d\,$ divides every denominator $\,n\,$ for $\,a$
Proof $\ \ (1\!\Rightarrow\! 2)\,\ \ \delta(a)\!=\!dk\Rightarrow a = c/(dk),\, (c,d)\!=\!1\,$ so $\,\delta(ka) = \delta(c/d) = d$
$(2\!\Rightarrow\!1)\ \ $ by $\,\color{#0a0}{\delta(ka)}\mid \delta(a),\,$ i.e. $\,\delta(a)a\in\Bbb Z\,\Rightarrow\,\delta(a)(ka)\in\Bbb Z\,\Rightarrow\, \color{#0a0}{d\ \mid\ } \delta(a)$
$(1\!\Rightarrow\! 3)\ \ \ na\in\Bbb Z\Rightarrow d\mid \delta(a)\mid n $
$(3\!\Rightarrow\! 1)\ \ \ \delta(a)a\in\Bbb Z\Rightarrow d\mid \delta(a),\,$ i.e. $\,d\,$ divides all denoms, including least denom.
The above Theorems can be unified when one studies (annihilator) ideals and modules.