3. Find all positive integer solutions to $(x+1)^{y+1}+1=(x+2)^{z+1}$.
(1999, Taiwan Mathematical Olympiad)
Let $x+1=a, y+1=b, z+1=c$. Then $a, b, c \geqslant 2$ and $a, b, c \in \mathbf{N}$. The original equation becomes
$a^{b}+1=(a+1)^{c}$.
For equation (1), take $\bmod (a+1)$, then we have $(-1)^{3}+1 \equiv 0(\bmod (a+1))$.
Thus, $b$ must be odd.
In fact, if $b$ is even, then $a=1$ leads to a contradiction.
Equation (1) becomes
$$
(a+1)\left(a^{b-1}-a^{b-2}+\cdots-a+1\right)=(a+1)^{c},
$$

which simplifies to
$$
a^{b-1}-a^{b-2}+\cdots-a+1=(a+1)^{c-1}.
$$

Taking $\bmod (a+1)$ on both sides of equation (2), we get
$$
\begin{array}{l}
1-(-1)+\cdots-(-1)+1 \\
=b \equiv 0(\bmod (a+1)).
\end{array}
$$

Thus, $(a+1) \mid b$.
Since $b$ is odd, $a$ must be even.
From equation (1), we have
$$
\begin{aligned}
a^{b}= & (a+1)^{c}-1 \\
= & (a+1-1)\left[(a+1)^{c-1}+(a+1)^{c-2}\right. \\
& +\cdots+(a+1)+1],
\end{aligned}
$$

which simplifies to
$$
\begin{aligned}
a^{b-1}= & (a+1)^{c-1}+(a+1)^{c-2}+\cdots \\
& +(a+1)+1.
\end{aligned}
$$

Taking $\bmod a$ on both sides of equation (3), we get
$$
0 \equiv 1+1+\cdots+1=c(\bmod a).
$$

Thus, $a \mid c$. Therefore, $c$ is even.
Let $a=2^{k} t(k \in \mathbf{N}, t$ be odd $), c=2 d(d \in \mathbf{N})$. Equation (1) becomes
$$
\begin{aligned}
2^{k b} t^{b} & =\left(2^{k} t\right)^{b}=\left(2^{k} t+1\right)^{2 d}-1 \\
& =\left[\left(2^{k} t+1\right)^{d}+1\right]\left[\left(2^{k} t+1\right)^{d}-1\right].
\end{aligned}
$$

Since $\left(\left(2^{k} t+1\right)^{d}+1,\left(2^{k} t+1\right)^{d}-1\right)=2$, there are only two cases:
(i) $\left(2^{k} t+1\right)^{d}+1=2 u^{b},\left(2^{k} t+1\right)^{d}-1=$ $2^{k-1} v^{b}$, where $2 \mid u, 2 \nmid v, (u, v)=1, u v=t$.
In this case, we have
$$
\begin{array}{l}
2^{k d-1} t^{d}+\mathrm{C}_{d}^{1} 2^{k(d-1)-1} t^{d-1}+\cdots+\mathrm{C}_{d}^{d-1} 2^{k-1} t+1 \\
=u^{b},
\end{array}
$$

Thus, $u \mid 1$, i.e., $u=1$.
Therefore, $\left(2^{k} t+1\right)^{d}=1$ is impossible.
(ii) $\left(2^{k} t+1\right)^{d}+1=2^{b-1} v^{b},\left(2^{k} t+1\right)^{d}-1=$ $2 u^{b}$, where $u, v$ satisfy the same conditions as in (i).
In this case, $2^{k} t \mid 2 u^{b}$, i.e., $2^{k} u v \mid 2 u^{b}$, so $k=1, v=1$. Thus, $2=2^{b-1}-2 u^{b}$, i.e., $u^{b}+1=2^{b-2}$.
Hence, $u=1, b=3, a=2, c=2$.
In conclusion, the original equation has a unique solution
$$
x=1, y=2, z=1.
$$