Certain properties of numbers such that $n \mid 2^n+1$

Here is a proof of part 1. Assume, $$ n_1 \mid 2^{n_1}+1 \ \text{and} \ n_2 \mid 2^{n_2}+1. $$ Denote, $d=gcd(n_1,n_2)$. We have, $$ 2^{n_1}\equiv -1 \pmod{d} \ \text{and} \ 2^{n_2}\equiv -1 \pmod{d}. $$ Now, from Bézout's identity, there exists $a,b\in\mathbb{Z}$ such that $d=an_1+bn_2$. Since, all of $d,n_1,n_2$ are odd, exactly one of $a,b$ is odd and the other one is even. Thus, $a+b$ is necessarily odd. Now, $$ 2^d \equiv 2^{an_1}2^{bn_2}\equiv (-1)^a(-1)^b \equiv (-1)^{a+b}\equiv -1 \pmod{d}, $$ proving that $(n_1,n_2) \mid 2^{(n_1,n_2)}+1$.

For the story involving least common multiple, let us assume that the set of all prime divisors of $n_1$ and $n_2$ is $\{p_1,p_2,\dots,p_k\}$ (after taking unions of all prime divisors of $n_1$ and $n_2$). Hence, by unique factorization theorem, there exists nonnegative integers $\{\alpha_1,\dots,\alpha_k\}$ and $\{\beta_1,\dots,\beta_k\}$ such that: $$ n_1 = p_1^{\alpha_1}\cdots p_k^{\alpha_k} \ \text{and} \ n_2 = p_1^{\beta_1}\cdots p_k^{\beta_k}. $$ Note that some of these numbers can very well be 0. Now, since $$ 2^{n_1}+1 \mid 2^{[n_1,n_2]}+1 \ \text{and} \ 2^{n_2}+1 \mid 2^{[n_1,n_2]}+1 $$, we have $$ n_1 \mid 2^{[n_1,n_2]}+1 \ \text{and} \ n_2 \mid 2^{[n_1,n_2]}+1. $$ Therefore, in the prime factorization of $2^{[n_1,n_2]}+1$, if the corresponding weights for $p_1,\dots,p_k$ are precisely $\theta_1,\dots,\theta_k$, we have $$ \theta_i \geq \alpha_i \ \text{and} \ \theta_i\geq\beta_i,\forall i = 1,2,\dots,k \implies \theta_i \geq \max\{\alpha_i,\beta_i\},\forall i. $$ Since the exponent of the prime $p_k$ in the factorization of $[n_1,n_2]$ is precisely $\max\{\alpha_i,\beta_i\}$, we are done.

Next, we will proceed into part 2. Assuming same notation for the prime decomposition of $n_1$ and $n_2$, we shall prove that $$ p_i^{\alpha_i+\beta_i}\mid 2^{n_1n_2}+1, \forall i. $$ Here's how to do it. Start with $p_1$, and assume that $\alpha_1 \geq \beta_1$ (you can do the exact same argument for the other case by swapping only one step below).

$$2^{n_1n_2}+1 = (2^{n_1}+1)\underbrace{((2^{n_1})^{n_2-1}-(2^{n_1})^{n_2-2}+\dots + 1)}_{\triangleq (*)}.$$ Clearly, since $n_1 \mid 2^{n_1}+1$, $p_1 ^{\alpha_1} \mid 2^{n_1}+1$. We will now prove that the second term in the factorization above is divisible by $p_1^{\beta_1}$, and by similar logic for the rest of the $p_2,\dots,p_k$, we will be done.

By our assumption, $\beta_1 \leq \alpha_1$, we have that $p_1^{\beta_1}\mid p_1^{\alpha_1}\mid 2^{n_1}+1$. Hence, $$ 2^{n_1}\equiv -1 \pmod{p_1^{\beta_1}}. $$ Now, we will compute $(*)$ modulo $p_1^{\beta_1}$. It is easy to see that $$ (*) \equiv \underbrace{1 + 1 + \dots + 1}_{n_2 \ \text{times}}\equiv n_2 \equiv 0 \pmod{p_1^{\beta_1}} $$ hence, we are done.

Had it been the case $\beta_1 \geq \alpha_1$, we could have used the equivalent factorization: $$ (2^{n_2}+1)((2^{n_2})^{n_1-1}-(2^{n_2})^{n_1-2}+\dots + 1) $$ to arrive at the result. Last step is to execute the exact same steps for $p_2,p_3,\dots,p_k$, in an analogous way, and conclude via the fact that the numbers $z_i \triangleq p_i^{\alpha_i+\beta_i}$ are mutually coprime, namely $(z_i,z_j) = 1$ if $i \neq j$. Since we have shown that each of $z_i$ divides $2^{n_1n_2}+1$ and they are all coprime, it must be the case that their product also divides $2^{n_1n_2}+1$. $\Box$