Prove that if $t$ is a natural number then there exists a natural number $n > 1$ such that $(n, t) = 1$ and none of the numbers $n + t, n^2 + t, n^3 + t…$ are perfect powers.
There is a solution posted at AOPS, by considering $n = t(t+1)^2 + 1$. Two cases are discussed: $t + 1$ is not a perfect power and $t + 1$ is a perfect power.
I have no problem understand the first case: if $t + 1$ is not a perfect power, then $n^k + t$ can be expressed in the form of $(t + 1)(b(t + 1) + 1)$, since $(t + 1)\nmid(b(t + 1) + 1)$, $(t + 1)$ has to be perfect power if $n^k + t$ is a perfect power. But this is contradiction.
But I got lost when trying to understand the second case, especially from "Consider $n^k+t=c^d$ so by proof...", can anyone help?