1. Define Electric Integers:
Let $p_1 = 2 < p_2 = 3 < \cdots$ denote the primes in increasing order. Call a positive integer $n$ electric if there exists $k \in \mathbb{N}$ such that
$$n = p_1^{\alpha_1} \cdots p_k^{\alpha_k},$$
where $\alpha_i > 0$, and $p_i - 1 \mid n$ for all $i$.

2. Property of Electric Integers:
Note that electric integers have the following property: If $m \equiv m' \pmod{n}$, and $m, m' \geq \alpha_1$, then
$$2^m + m \equiv 2^{m'} + m' \pmod{n}.$$
Furthermore, if $n$ is electric, then $n/p_k^{\alpha_k}$ is also electric.

3. Induction Hypothesis:
We claim that for any electric integer $n$, there exists a positive integer $m \geq \alpha_1$ with $n \mid 2^m + m$. We proceed by induction on $k$, the number of distinct prime factors of $n$.

4. Base Case:
For $k = 1$, $n = p_1^{\alpha_1}$. We can choose $m = 2^{\alpha_1}$, which trivially satisfies $n \mid 2^m + m$.

5. Inductive Step:
Assume the statement is true for $k$. Let $n = p_1^{\alpha_1} \cdots p_k^{\alpha_k}$ be an electric integer. Write $n' = n/p_k^{\alpha_k}$, which is also electric by definition. By the induction hypothesis, there exists $m \geq \alpha_1$ such that $n' \mid 2^m + m$.

6. **Constructing $m_{\alpha}$:**
We claim that for all $\alpha \geq 0$, we can find $m_{\alpha} \geq \alpha_1$ such that
$$n'p_k^{\alpha} \mid 2^{m_{\alpha}} + m_{\alpha}, \quad m_{\alpha+1} \equiv m_{\alpha} \pmod{n'p_k^{\alpha}}.$$
We proceed by induction on $\alpha$.

7. **Base Case for $\alpha$:**
For $\alpha = 0$, set $m_0$ to be the value that worked for $n'$.

8. **Inductive Step for $\alpha$:**
Assume it is true for $\alpha$. To show it is true for $\alpha + 1$, set $m_{\alpha+1} = m_{\alpha} + c n' p_k^{\alpha}$ for some choice of $c \geq 0$. Note that $n' p_k^{\alpha}$ is electric, so
$$n' p_k^{\alpha} \mid 2^{m_{\alpha+1}} + m_{\alpha+1}.$$
Furthermore, since $m_{\alpha+1} \equiv m_{\alpha} \pmod{(p_k-1)p_k^{\alpha}}$, we have $2^{m_{\alpha+1}} \equiv 2^{m_{\alpha}} \pmod{p_k^{\alpha+1}}$.

9. **Choosing $c$:**
Pick $c$ such that
$$-c n' p_k^{\alpha} \equiv 2^{m_{\alpha}} + m_{\alpha} \pmod{p_k^{\alpha+1}},$$
or
$$c n' \equiv -\frac{2^{m_{\alpha}} + m_{\alpha}}{p_k^{\alpha}} \pmod{p_k}.$$
This is possible because $p_k^{\alpha} \mid 2^{m_{\alpha}} + m_{\alpha}$ and $\gcd(n', p_k) = 1$.

10. Conclusion:
Since $m_{\alpha+1} \equiv m_{\alpha} \pmod{n'}$, we have
$$2^{m_{\alpha+1}} + m_{\alpha+1} \equiv 0 \pmod{n'},$$
and also
$$2^{m_{\alpha+1}} + m_{\alpha+1} \equiv 2^{m_{\alpha}} + m_{\alpha} - c n' p_k^{\alpha} \equiv 0 \pmod{p_k^{\alpha+1}}.$$
This implies that $2^{m_{\alpha+1}} + m_{\alpha+1}$ is divisible by $n' p_k^{\alpha+1}$, completing the induction. Note that $m_{\alpha+1} \geq m_{\alpha} \geq \alpha_1$.

11. Final Step:
Thus, if $n$ is electric, we can find $m$ with $2^m + m \equiv 0 \pmod{n}$. Now, take $n = p_1^{\beta_1} \cdots p_k^{\beta_k}$. We set
$$\gamma_i = \max(1, \beta_i, \nu_{p_i}(p_{i+1}-1), \nu_{p_i}(p_{i+2}-1), \ldots, \nu_{p_i}(p_{k}-1))$$
for $1 \leq i \leq k$. Then $N = p_1^{\gamma_1} \cdots p_k^{\gamma_k}$ is electric, and so we can find $m$ such that
$$n \mid N \mid 2^m + m$$
as desired.

$\blacksquare$