$$2^5 \times 9^2 =2592$$ I am trying to find any other number in this pattern.
That is find natural numbers $a$ , $b$ , $c$ and $d$ such that $$a^b \times c^d = \overline{abcd} $$
We have $$a^b \times c^d = \overline{cd} + 100\overline{ab} $$
So $$a^b \times c^d - \overline{cd} = 100\overline{ab} $$
LHS is a multiple of $100$ .Any help from here will be greatly appreciated
The 2592 puzzle apparently originated with Henry Ernest Dudeney's 1917 book Amusements in Mathematics where it is given as puzzle 115, "A PRINTER'S ERROR":
In a certain article a printer had to set up the figures $5^4\times2^3,$ which of course meant that the fourth power of $5\ (625)$ is to be multiplied by the cube of $2\ (8),$ the product of which is $5,000.$ But he printed $5^4\times2^3$ as $5423,$ which is not correct. Can you place four digits in the manner shown, so that it will be equally correct if the printer sets it up aright, or makes the same blunder?
[. . . .]
The answer is that $2^5\times9^2$ is the same as $2592,$ and this is the only possible solution to the puzzle.
It was apparently rediscovered fifteen years later and published in the American Mathematical Monthly, vol. 40, December 1933, p. 607, as problem E69, proposed by Raphael Robinson, in the following form:
Instead of a product of powers, $a^bc^d,$ a printer accidentally prints the four digit number, $abcd.$ The value is however the same. Find the number and show that it is unique.
A solution by C. W. Trigg was published in vol. 41, May 1934, p. 332; the problem was also solved by Florence E. Allen, W. E. Buker, E. P. Starke, Simon Vatriquant, and the proposer.
Asking Mathematica to do this
Do[If[a^b*c^d==a*1000+b*100+c*10+d,Print[{a,b,c,d}];,],{a,1,9},{b,1,9},{c,1,9},{d,1,9}] returns unique $2592$ you already have.
You said you are looking for solutions with natural numbers.
If c = 1 or d = 0 then $c^d = 1$, so you are left with $a^b = ab1d$ or $a^b = abc0$. You might be able to find solutions here, since all you need is that $a^b$ starts with ab, followed by the digit 1 and any other digits, or that $a^b$ starts with ab, followed by any other digits, and ending in 0.
$a^b$ can be arbitrary large, so you have a large search space. Since $a^b$ must start with a, $a^{b-1}$ must be just slightly larger than a power of 10, so a would be just a bit larger than $10^{k/(b-1)}$.
For example, $316,227,766,017^3 ≈ 316,227,766,017.324 \times 10^{23}$, so if we let a = 316,227,766,017 and b = 3, then $a^b$ starts with $ab$. Here the next digit is 2, but I would be quite sure that there will be cases where the next digit is 1; they will be a bit hard to find because you will need to perform arithmetic with a few hundred or more decimal digits.
You'd have to check it, but I think you get a solution with a = 1,930,697,729 and b = 8, c = 1 and d some large number with the remaining digits. $a^b$ starts with 1,930,697,729,817. $a = 10^9 \times 10^{2/7}$, rounded up to the nearest integer.
BTW I didn't use Mathematica but a spreadsheet :-)
