I interpret the problem as follows:

Let $Y=X_1+X_2$, where $X_1$ and $X_2$ are independent random variables (r.v.'s). We want to maximize the entropy of $Y$ with respect $X_2$ given a finite variance of $Y$, with the r.v. $X_1$ fixed. Is it true that the entropy of $Y$ is maximized if and only if $X_2$ has a normal distribution?

The answer to this question is no. Indeed, suppose that $P(X_1=1)=P(X_1=-1)=1/2$ and $Var\,Y=1+s^2$ for some real $s>0$, so that $Var X_2=s^2$. Suppose that $X_2$ is normally distributed, with a pdf $g$. Since the entropy of a pdf is invariant with respect to shifts, without loss of generality $X_2\sim N(0,s^2)$.

Let now $h$ be another pdf with mean $0$ and variance $s^2$. For $t\in[0,1]$, let
$$f_t:=(1-t)g+th,$$
which is the pdf of a r.v. $X_{2,t}$ with variance $s^2$ such that $X_{2,t}$ is independent of $X_1$. So, the pdf $p_t$ of $Y_t:=X_1+X_{2,t}$ (with the same variance, $1+s^2$, as $Y$) is given by the formula
$$p_t(y)=\tfrac12\,f_t(y+1)+\tfrac12\,f_t(y-1)
$$
for real $y$. The entropy of $Y_t$ is
$$H(t):=-\int_{\mathbb R}p_t(y)\ln p_t(y)\, dy,
$$
and $H(0)$ is the entropy of $Y$.
Next,
$$H'(0)=-\int_{\mathbb R}[1+\ln p_0(y)]\,\frac\partial{\partial t}\,p_t(y)\Big|_{t=0}\, dy \\
=\int_{\mathbb R}\left[\frac{g(y+1)+g(y-1)}2-\frac{h(y+1)+h(y-1)}2\right]\,
\ln\frac{g(y+1)+g(y-1)}2\, dy.
$$
To disprove the conjecture, it is enough to find a suitable pdf $h$ such that $H'(0)>0$. In view of Chebyshev's integral inequality, it is then natural to choose $h$ which is almost mutually singular with $g$, so that $h$ is small where $g$ is large, and vice versa. With this in mind, let $h$ be the half-and-half mixture of the pdf's of $N(s\sqrt{1-u^2},u^2s^2)$ and $N(-s\sqrt{1-u^2},u^2s^2)$ with $u\in(0,1)$, so that the mean and variance of $h$ are indeed $0$ and $s^2$, respectively. Then
$H'(0)=0.05121\ldots>0$ if $s=1/2$ and $u=1/20$, and we are done.

Here are graphs of the almost mutually singular $g$ (blue) and $h$ (yellow):