Seventh question: Find all pairs of positive integers $(a, n)$ such that $a \geq n$, and $(a+1)^{n}+a-1$ is a power of 2.
Question 7:
Solution: If $n=1$, then $(a+1)^{n}+a-1=2a$, so $a$ is a power of 2. When $(a, n)=\left(2^{k}, 1\right), k \in \mathbb{N}$, it is a solution that satisfies the condition.
Assume $a \geq n \geq 2$ below.
Let $(a+1)^{n}+a-1=2^{m}$. Since $a \mid\left[(a+1)^{n}+a-1\right]$, it follows that $a \mid 2^{m}$. Let $a=2^{k}, k \in \mathbb{Z}^{+}$, then $\left(2^{k}+1\right)^{n}+2^{k}-1=2^{m}$.

Since $2^{m}=\left(2^{k}+1\right)^{n}+2^{k}-1 \geqslant\left(2^{k}+1\right)^{2}+2^{k}-1=2^{2k}+2^{k+1}+2^{k}>2^{2k}$, it follows that $2^{2k} \mid 2^{m}$. According to the binomial theorem, we have:
$$
0 \equiv\left(2^{k}+1\right)^{n}+2^{k}-1=\sum_{t=0}^{n} C_{n}^{t} 2^{kt}+2^{k}-1 \equiv(n+1) 2^{k} \pmod{2^{2k}}
$$

Thus, $2^{k} \mid (n+1)$. Combining $2^{k}=a \geq n$, we get $n=2^{k}-1$. Since $n \geq 2$, it follows that $k \geq 2$.
When $k=2$, $a=4, n=3$, at this time $(a+1)^{n}+a-1=2^{7}$, which satisfies the condition.
When $k \geq 3$, $n \geq 7$. Then $2^{m}=\left(2^{k}+1\right)^{n}+2^{k}-1 \geq\left(2^{k}+1\right)^{7}+2^{k}-1>2^{7k}$, so $2^{7k} \mid 2^{m}$, hence $2^{3k} \mid 2^{m}$. According to the binomial theorem, we have:
$$
\begin{array}{l}
0 \equiv\left(2^{k}+1\right)^{n}+2^{k}-1=\sum_{t=0}^{n} C_{n}^{t} 2^{kt}+2^{k}-1 \equiv \frac{n(n-1) 2^{2k}}{2}+(n+1) 2^{k} \\
\equiv \frac{\left(2^{k}-1\right)\left(2^{k}-2\right) 2^{2k}}{2}+2^{2k} \equiv\left(2^{k}-1\right)\left(2^{k-1}-1\right) 2^{2k}+2^{2k} \pmod{2^{3k}} \\
\left.\left.2^{k}\right]\left(2^{k}-1\right)\left(2^{k-1}-1\right)+1\right] .
\end{array}
$$

Thus, $2^{k} \mid\left[\left(2^{k}-1\right)\left(2^{k-1}-1\right)+1\right]$.
In fact, $\left(2^{k}-1\right)\left(2^{k-1}-1\right)+1 \equiv-2^{k-1}+2 \equiv 0 \pmod{2^{k}}$, which leads to a contradiction.
In summary, the solutions that satisfy the condition are $(a, n)=(4,3)$ or $\left(2^{k}, 1\right), k \in \mathbb{N}$.