"Finding all positive integer solutions to the equation $x^{y}=y^{x}+1$ is a problem I studied over twenty years ago. At that time, I provided a solution, but there was actually an issue with one step in the process. It was pointed out by readers after I included this problem in the 2015 Special Issue of *High School Mathematics* in the *National High School Mathematics League Simulation Questions (13)*. Only then did I reconsider the solution.
In August 2017, a student from Tsinghua University, Chen Jiahao, informed me that this problem appears on page 241 of reference [2].
This article presents two new solutions different from those in reference [2]."

保留了原文的换行和格式。
```
    Solution 1 (1) If }x\text{ is even; }y\text{ is odd}
    Let }x=2m(m\in\mathbb{Z}^+)\text{. Then
    }ym=(2m)^y-
    # y^2m-1=(2m)^y-2
    # (ym+1)(y^m-1)=(2m)^y-
    and }ym+1, y^m-1\text{ are two consecutive even numbers, so
    4|((2m)^y -2).
    Therefore, }y=1, \text{ substituting into the original equation gives }x=2\text{, i.e., (2,1)
is a solution to the original equation.
    (2) If }x\text{ is odd, }y\text{ is even.
    }x^{2m}=(2^\alpha m)^x+1
    * x^{2m}-1=(2^\alpha m)^x
    Note that, }2^{2m}-1=(x^m)^2-1
    = (x^{2m} - 1) \prod_{i=1}^{\infty-1} (x^{2m} + 1)
    Also, by }x\equiv1(\operatorname{mod}2)\text{, we have
    }x^m\equiv1(\operatorname{mod}2
    Thus, the equation holds when }2^{\alpha+1}|(x^{2m}-1)\text{.
}2^{\alpha+1}|(x^2-1)\text{, which must satisfy
    However, }\alpha\in\mathbb{Z}^+\text{, when }x\geqslant4\text{,
    2
it contradicts equation (1);
    When }x=1\text{, equation (1) is obviously not true
    Also, }3
    = (y+1)(y^2-y+1)
    =(y+1)((y+1)(y-2)+3),
y
    y+1=3 \text{ or }y^2-y+1=3\text{.
    In summary, }y=2\text{.
    Therefore, when }x\text{ is odd, there is only one solution (3,2)
    In conclusion, the original equation has only two solutions (2,1),(3,2)
    Solution 2 When }x=1,2,3\text{, the solution is the same as Solution 1
    Next, we prove: }x\geqslant4\text{, then }y>x
    Proof by contradiction.
    Clearly, }x\neq y\text{, so }y\neq x\text{.
    Hence, }y>x\geqslant4\text{.
    In the indeterminate equation }y^y = x^x + 1\text{, let
    }y=x+\alpha(\alpha\in\mathbb{Z}^+)\text{.
    Then }x^{x+\alpha}-(x+\alpha)^x=
    Define the function }f(x)=x^{x+\alpha}-(x+\alpha)^x (x\geqslant1)\text{.
    Therefore, }f'(x)=(\ln x+\frac{x+\alpha}{x})x^{x+\alpha}
            -(\ln(x+\alpha)+\frac{x}{x+\alpha})(x+\alpha)^x
    Let }g(x) = (\ln x + \frac{x + \alpha}{x}) - (\ln(x + \alpha) + \frac{x}{x + \alpha})
    = \ln\frac{x}{x+\alpha}+\frac{x+\alpha}{x}-\frac{x}{x+\alpha}\text{.
    Let }t=\frac{x+\alpha}{x}>1\text{. Then
    }h(t) = \ln\frac{1}{t} + t - \frac{1}{t}\text{,
    }h'(t)=1+\frac{1}{t^2}-\frac{1}{t}>0\text{.
    Therefore, }h(t)>h(1)=0\text{.
    }g(x)>0\text{.
    } \ln x+\frac{x+\alpha}{x}>\ln(x+\alpha)+\frac{x}{x+\alpha}>0\text{.
    Since }x^{x+\alpha}-(x+\alpha)^x=1\text{,
    }x^{x+\alpha}>0, (x+\alpha)^x>0\text{,
    Therefore, }f'(x)>0\text{.
    When }x\geqslant4\text{,
    }f(x)\geqslant f(4)=4^{4+\alpha}-(4+\alpha)^4\text{,
    }=(4+\alpha)^4(\frac{4^{4+\alpha}}{(4+\alpha)^4}-1)(\alpha\geqslant1)\text{.
    Let }q(\alpha)=\ln\frac{4^{4+\alpha}}{(4+\alpha)^4}
    = (4+\alpha)\ln 4-4\ln(4+\alpha)\text{,
    which is an increasing function, so
    }q'(\alpha)\geqslant q'(1)=\ln 4-\frac{4}{5}>0\text{,
    which is also an increasing function.
    Also, }r(\alpha)=\frac{4^{4+\alpha}}{(4+\alpha)^4}
    }r(1)=\frac{4^5}{5^4}-1=\frac{1024}{625}-1>0\text{,
    which is an increasing function, i.e., when }x\geqslant4\text{,
    }p(1)=4^5-5^4=399\text{.
    It is evident that when }x\geqslant4\text{, the original equation has no solution.
```