Notice that if $(x-1,x)$ is such a pair then $(4x(x-1),(2x-1)^2)$ is also such a pair. Starting at $(8,9)$ we can generate infinitely many such pairs.

We have $(2^3, 3^2)$ as a pair, and taking $4k(k - 1), (2k - 1)^2$ clearly works. edit: sniped

A different solution: It suffices to note that the Pell equation $x^2-8y^2=1$ has infinitely many solutions as then both $x^2$ and $x^2-1=8y^2$ are powerful. But this is well-known.

Notice that $8,9$ is a solution and then you can go from $(x, x+1)$ to $(4x^2 + 4x, 4x^2 + 4x + 1) = (4x(x+1), (2x+1)^2) $, which is clearly a different but working solution.