7. Find all positive integer solutions $(x, r, p, n)$ of the equation $x^{\prime}-1=p^{n}$ that satisfy the following two conditions:
(1) $p$ is a prime; (2) $r \geqslant 2, n \geqslant 2$. (4th test, 2nd question, an example related to Catalan's conjecture)
If $x=2, 2^{r}-1$ is an odd number $(r \geqslant 2)$, then $p$ is an odd number. If $n$ is even, then $p^{n}+1 \equiv 2 \bmod 4$, thus $2^{r} \equiv 2 \bmod 4, r=1$, no solution. If $n$ is odd, then $2^{r}=p^{n}+1=(p+1)\left(p^{n-1}-p^{n-2}+\cdots-p+1\right)$, hence we can set $p+1=2^{t}(t \in N)$. Then, $p^{n}+1=\left(2^{i}-1\right)^{n}+1=\sum_{i=0}^{n-1}(-1)^{i} C_{n}^{i} - \left(2^{t}\right)^{n-i}=n 2^{t}+2^{z x} M$, where $M$ is an integer. Thus, $p^{n}+1$ can be divided by $2^{t}$, but not by $2^{t+1}$, and since $p^{n}+1=2^{r}$, we have $r=t, p^{n-1}-p^{n-2}+\cdots-p+1=1$, and since $n \geqslant 2, n$ is odd, this equation cannot hold. Therefore, when $x=2$, the original equation has no solution.

If $x \geqslant 3, p^{n}=(x-1)\left(x^{r-1}+x^{r-2}+\cdots+x+1\right)$, then $x-1$ is a power of $p$, and since $x-1 \geqslant 2$, we can set $x-1=p^{m}(m \in N)$. Thus, $x \equiv 1 \bmod p$ and $r \equiv x^{r-1}+x^{r-2}+\cdots+x+1 \equiv 0 \bmod p$.

If $p=2, r$ is even, let $r=2 r_{1}\left(r_{1} \in N\right)$. Then, $2^{n}=x^{r}-1=\left(x^{r_{1}}-1\right)\left(x^{r_{1}}+1\right)$, and the difference between $x^{r_{1}}+1$ and $x^{r_{1}}-1$ is $2$, and both $x^{r_{1}}+1$ and $x^{r_{1}}-1$ are powers of $2$, so it must be that $x^{r_{1}}-1=3$, i.e., $x=3, r_{1}=1$, thus we have the solution $x=3, r=2, p=2, n=3$.

If $p \geqslant 3$, then $x^{r}-1=\left(1+p^{m}\right)^{r}-1=r p^{m}+\sum_{i=2}^{r} C_{r}^{i}\left(p^{m}\right)^{i}$, and each term in the sum $C_{r}^{i}\left(p^{m}\right)^{i}=C_{r-1}^{i-1} r p^{m} - \frac{\left(p^{m}\right)^{i-1}}{i}$. Since $2<\left(p^{m}\right)^{2-1}$, if we set $k<\left(p^{m}\right)^{k-1}$, where $k \geqslant 2$ is a positive integer, then $k+1<2 k<2\left(p^{m}\right)^{k-1}<\left(p^{m}\right)^{k}$. The above induction proves that for any positive integer $i \geqslant 2$, we have $i<\left(p^{m}\right)^{i-1}$. Since $C_{r-1}^{i-1}$ is an integer, each term $C_{r}^{i}\left(p^{m}\right)^{i}$ contains a higher power of $p$ than $r p^{m}$, which contains a power of $p$ of $\alpha (\alpha \geqslant m)$. Thus, $x^{r}-1$ can be divided by $p^{\alpha}$, but not by $p^{\alpha+1}$, and since $x^{r}-1=p^{\alpha}$, this leads to a contradiction.