Find the pairs!

The index of $3$ in $2018!$ is $1000$ , and the index of $7$ in $2008!$ is $331$ . Suppose that positive integers $x,y$ satisfy the given equation.

Since $3$ divides both $2008!$ and $21^y$ , it must divide $x^{2008}$ . Thus $3$ divides $x$ , and hence $3^{2008}$ divides $x^{2008}$ , so that $3^{2008}$ divides $x^{2008} = 21^y - 2008!$ . Since the index of $3$ in $2008!$ is $1000$ , the index of $3$ in $21^y$ must also be $1000$ (since otherwise the index of $3$ in $x^{2008}$ will be at most $1000$ ). Thus we deduce that $y = 1000$ .

By considering divisibility by $7$ , we deduce similarly that $y = 331$ . Thus there are no positive integer solutions to this equation.

Naturally, no pairs of $( x, y )$ satisfies that equation.