if $a,b\in \mathbb{N}\: ,\: a^{x}=b^{y}\Rightarrow \exists \: t\in \mathbb{N}:\: a=t^{s}\: ,\: b=t^{k}$ proof$a=\prod_{i=1}^{l}p_{i}^{\alpha _{i}}\prod_{j=1}^{n}q_{j}^{\gamma _{j}}\: \: ,\: \: b=\prod_{i=1}^{l}p^{\beta _{i}}_{i}\prod_{j=1}^{m}r^{\theta _{j}}_{j}$ it is easy to see that : $m=n=0$ so we have : $a=\prod_{i=1}^{l}p_{i}^{\alpha _{i}}\: \: ,\: \: b=\prod_{i=1}^{l}p^{\beta _{i}}_{i}$ now we have : $x\alpha _{i}=y\beta _{i}\: \: \overset{x=dp\: ,\: y=dq}{\rightarrow}\: \frac{\alpha _{i}}{\beta _{i}}=\frac{q}{p}$ so $q\mid \alpha _{i}\: ,\: p\mid \beta _{i}$ now let : $t=\prod_{i=1}^{l}p_{i}^{\frac{\alpha _{i}}{q}}=\prod_{i=1}^{l}p_{i}^{\frac{\beta _{i}}{p}}$ so we have : $a=t^{q}\: \:, b=t^{p}$

The only solutions are \[(x,y) = (1,1), (4,2), (5,25), (8,32), (27,81)\] Set \(x=da\) and \(y=db\) where \(d=\gcd(x,y)\) and \(\gcd(a,b) = 1\). Then the equation becomes \[a^{a+b} = d^{2a - b} b^{3a}\] Case 1: Suppose \(b\geq 2a\). Then we have \[b^{3a} = d^{b-2a} a^{a+b}\]As \(\gcd(a,b) = 1\), we must have \(a=1\). So we get \[b^3 = d^{b-2}\] Now we can manually check the cases \(d\in\{1,2,3,4\}\) to conclude that \(d\geq 5\). Therefore, \[b^3 = d^{b-2} \geq 5^{b-2} \implies b\leq 5\]so we check the cases \(b\in\{2,3,4,5\}\) to get the three solutions \((x,y)=(27,81), (8,32), (5,25)\) Case 2: Now assume that \(2a > b\). Then as before we must have \(b=1\), so we have \[a^{a+1} = d^{2a-1}\]Note that for any prime \(p\), we have \[2a-1 \mid (a+1)\nu_p(a)\]So we consider two more subcases: Subcase 1: Assume that \(a+1\) is not a multiple of \(3\), so that \(\gcd(a+1,2a - 1) = 1\). Then, \[2a - 1 \mid \nu_p (a) \implies a \geq p^{2a-1} \geq 2^{2a-1}\]which is impossible Subcase 2: The only other possibility is that \(a+1\) is a multiple of \(3\). In this case, we find that \(\gcd(a+1,2a-1) = 3\), so that \[\frac{2a-1}{3}\mid \nu_p(a) \implies a \geq p^{\frac{2a-1}{3}} \geq 2^{\frac{2a-1}{3}}\]which is only possible if \(a=2\). This gives the solution \((x,y) = (4,2)\). Finally, note that \((1,1)\) is also a solution, so we are done.