$ x=(1+\frac{1}{n})^{n}$ , $ y=(1+\frac{1}{n})^{n+1}$ $ n$ positive integer number . $ x^{y}=[(1+\frac{1}{n})^{n}]^{(1+\frac{1}{n})^{n+1}}=(\frac{n+1}{n})^{\frac{(n+1)^{n+1}}{n^{n}}}$ $ y^{x}=[(1+\frac{1}{n})^{n+1}]^{(1+\frac{1}{n})^{n}}=(\frac{n+1}{n})^{\frac{(n+1)^{n+1}}{n^{n}}}$

Nice construction, but how do you know they are the only ones?

Peter wrote: Nice construction, but how do you know they are the only ones? You can find this problem in "American Mathematical Monthly".If someone wants,i could post a solution.

We surely do! Also, can you please post the full reference? Then I can include it.

I'm sorry for my late reply. Problems author: Oystein Ore Problem: Find all pairs of positive rational numbers $ a,b$,such that $ a^b=b^a$ Solution: Obviously $ a=b$,is the solution. Now WLOG assume that $ b>a$. Then let $ b=ra$,where $ r>1$,$ r\in\mathbb{Q_+}$,and $ a^{r-1}=r$,or $ a=r^{\frac{1}{r-1}}$. Let $ \frac{1}{r-1}=\frac{u}{v}$,where $ u,v\in\mathbb{N}$ are relatively prime numbers. Since $ a$ and $ r$ are rational numbers,then $ r=\frac{u+v}{u}$ should be $ v$-th power of a rational number. But then numbers $ u$ and $ u+v$,as relatively primes,must be $ v$-th power of an integer number. It is well-known and easy to prove,that an absolute value of the diffrence between the two $ v$-th powers of a natural numbers,is greater than $ v$,if $ v>1$. Hence $ v=1$,and $ \frac{1}{r-1}=u$ must be an integer number. So $ a=(\frac{u+1}{u})^u$ and $ b=(\frac{u+1}{u})^{u+1}$

1) There is an other interesting method by Sierpinski: We are going to find all the solutions of \[ (1)\hfill \qquad \qquad \qquad x^{y} = y^{x}, \] in positive rational numbers, x,y with $ x \ne y$. Suppose that x,y is such a solution and x>y.Then $ r = ( 1 + \frac{1}{r} )x$ is a positive rational number and $ y = (1+\frac{1}{r})x$. Therefore ${ x^{y} = x ^ (1+\frac{1}{r}) x}$ and, since $ x^{y}=y^{x}$ we have also $ x^{(1+\frac{1}{r})}=y^{x}$ which proves that $ x^{1+\frac{1}{r}}=y= (1+\frac{1}{r}) x$. Hence $ x^{\frac{1}{r}}= 1+\frac{1}{r}$ and $ x=(1+\frac{1}{r} )^{r}$, $ y=(1+\frac{1}{r} )^{r+1}$. Let $ r=\frac{n}{m}$, with $ (m,n)=1$ and $ x=\frac{t}{s}$, with $ (t,s)=1$. Since $ x= ( 1+ \frac{1}{r})^{r}$ and we have $ \frac {m+n}{n})^{n\m}=\frac{t}{s}$ whence $ ( \frac{m+n}{n})^{n}=(\frac{t}{s})^{m}$. Each side of this equality is an irreductible fractions.Since $ (m,n)=1$ then $ (m+n,n)=1$ and since $ (t,s)=1$ then ( $ t^{m}$ , $ s^{m}$)=1 . Now, it follows that $ (m+n)^{n}=t^{m}$ and $ n^{n}=s^{m}$. By $ (m,n)=1$ we infer that there exist natural numbers $ k$ and $ \ell$ such that $ m+n =k^{m}$, $ n= \ell^{m}$ and $ = \ell^{n}$. Therefore $ m+\ell^{m}=k^{m}$. From this we deduce that $ k\geq \ell+1$. If $ m>1$ we would have $ k^{m} > ( \ell+1 ) ^{m} \geq \ell^{m}+ m \ell^{m-1}+1> \ell^{m}+m=k^{m}$ which is impossible. So $ m=1$, whence $ r=\frac{n}{m}=n$. The conclusion is that \[ (2)\hfill \qquad \qquad \qquad x= ( 1+ \frac{1}{n})^{n}, \quad y= ( 1+ \frac {1}{n})^{n+1} \] with n natural number.Conversely, x and y verify the equation $ x^{y}=y^{x}$. Therefore, all the solutions of the equation in the rational numbers are given by (2) with n natural number. It follows that n=1 is only value for which the equation has a solution in natural numbers $ (x,y)= (2,4)$. 2) There is an other possible solution but only for positive integers, using : Lemma: If $ a^{n}\mid b^{n}$ , then $ a\mid b$ for all $ a, b, n$, positive integers. Now let us suppose that in the equation $ x^{y}=y^{x}$ one has $ y>x$ .Let $ y =x+t$ (t>0).Then (1) can be written as:$ x^{x+t}=y^{x}$ or $ x^ {x} \cdot x^{t} =y^{x}$ which implies: $ x\mid y$. So, by Lemma $ (a=x, b=y, n=x)$ we must have $ x\mid y$. Let $ y=kx$, $ (k>1)$. Then the equation becomes $ x^{kx}=y{x}$ , yielding $ kx=x^{k}$ or $ x^{k-1}= k$. For $ k=2$ we get 2. For $ k\geq 3$ by induction we show that $ x^{k-1}$ >k ( for $ x\geq 2$), so the equation is impossible.Therefore $ k=2,x=2,y=4$.