Finiteness of counterexample follows from a paper On the zeros of sums of the Riemann zeta function

From p.1:

For fixed real number $\sigma_0$ define

$$ H(\sigma_0,s) = \zeta(\sigma_0 + s) + \zeta(\sigma_0 -s) \text { or } \zeta(\sigma_0 + s) - \zeta(\sigma_0 -s) $$

Your question is about $\sigma_0 = 0$.

From p. 3, Theorem 1:

2. If $\sigma_0 \le 0$, then all but finitely many complex zeros of $H(\sigma_0,s)$ are on $\Re(s)=0$.

This is unconditional.

A generalization:

1. If $\sigma_0 \le \frac12$, then all but finitely many complex zeros of $H(\sigma_0,s)$ are on $\Re(s)=0$ provided that $\zeta(s)$ has only finitely many complex zeros in $\Re(s) < \sigma_0$.

Related Theorem 3, p.4.

(1) Let $\sigma_0 < \frac12$ . Then, all zeros of $H(\sigma_0,s)$ in $|\Im(s)| \ge 100$ are on $\Re s = 0$ provided that $\zeta(s)$ in $\Re(s) < \sigma_0$ and $\Im(s) \ne 0$ has no zeros.

(2) The Riemann hypothesis holds if and only if for any $\sigma_0 <\frac12$ , all zeros of $H(\sigma_0,s)$ in $|\Im(s)| \ge 100$ are on $\Re(s) = 0$.

Finiteness of counterexamples follows from the paper On the zeros of sums of the Riemann zeta function, Haseo Ki

From p.1:

For fixed real number $\sigma_0$ define

$$ H(\sigma_0,s) = \zeta(\sigma_0 + s) + \zeta(\sigma_0 -s) \text { or } \zeta(\sigma_0 + s) - \zeta(\sigma_0 -s) $$

Your question is about $\sigma_0 = 0$.

From p. 3, Theorem 1:

2. If $\sigma_0 \le 0$, then all but finitely many complex zeros of $H(\sigma_0,s)$ are on $\Re(s)=0$.

This is unconditional.

A generalization:

1. If $\sigma_0 \le \frac12$, then all but finitely many complex zeros of $H(\sigma_0,s)$ are on $\Re(s)=0$ provided that $\zeta(s)$ has only finitely many complex zeros in $\Re(s) < \sigma_0$.

Related Theorem 3, p.4.

(1) Let $\sigma_0 < \frac12$ . Then, all zeros of $H(\sigma_0,s)$ in $|\Im(s)| \ge 100$ are on $\Re s = 0$ provided that $\zeta(s)$ in $\Re(s) < \sigma_0$ and $\Im(s) \ne 0$ has no zeros.

(2) The Riemann hypothesis holds if and only if for any $\sigma_0 <\frac12$ , all zeros of $H(\sigma_0,s)$ in $|\Im(s)| \ge 100$ are on $\Re(s) = 0$.