This question is close since a long time, but I would like to add my solution, that could interest readers.

First, we can assume d is odd.

Let's define the following sequence:

\begin{align}
a_0 &= 1\\
a_{n+1} &= 2^{b_n}a_n-d\\ &\quad\text{Where } {b_n} \text{ is the only integer } b_n \ge 1\text{ such that } 2^{b_n-1}a_n\leq d \lt 2^{b_n}a_n.
\end{align}

We show by induction that $a_n$ is well defined, odd and that $1 \leq a_n \leq d$.

Since $a_n$ exist and $1\leq a_n \leq d$, $a_{n+1}$ exist.

Since $1 \leq b_n$ and d odd, $a_{n+1}$ is odd.

Clearly we have $1 \leq a_{n+1}$. Since $2^{b_n-1}\leq \frac{d}{a_n} \lt 2^{b_n}$ we have $2^{b_n}-\frac{d}{a_n} \leq 2^{b_n-1}$ and so $a_{n+1}\leq d$.

We deduce that the sequence $(a_n)$ is periodic from a certain rank, at least two terms between $a_0,a_1,..,a_{\frac{d+1}{2}}$ must be equal ($a_n$ is odd so we have $\frac{d+1}{2}$ possibilities for $a_n$).

We can rewrite the sequence of the problem, eluding even terms and with a different initial condition, as:

\begin{align}
u_0 &\in 2\mathbb{N}+1\\
u_{n+1} &= \frac{u_n+d}{2^{v_2(u_n+d)}}\\ &\quad\text{Where } {v_2(n)} \text{ is the 2-adic valuation of n.}
\end{align}

Taking $u_0=a_n$, we have $u_k=a_{n-k} $ and so $u_n=a_0=1$.

Now, if $a_n$ do not take the value 1 for any $0 \lt n \leq \frac{d+1}{2}$, then it's period don't contain 1, and so taking $u_0=a_\frac{d+1}{2}$ we can't have $u_\frac{d+1}{2}=1$.

Then there exist $0 \lt n \leq \frac{d+1}{2}$ such that $a_n=1$.

Then taking $u_0=a_n=1$ we have $u_n=1$.

So, with $u_0=1$, there exist $0 \lt n \leq \frac{d+1}{2}$ such that $u_n=1$.