Find integer solution(s) to $(n-2)(p^2+n)^{14}=z^{13}$. Background. Just a spinoff from my second answer to this question, Does the equation $a^{2} + b^{7} + c^{13} + d^{14} = e^{15}$ have a solution in positive integers where I found parametric “almost answers” using $p^2+1$ and $p^2+2$. I’m looking at the very remote possibility that a full answer might follow the pattern, so the problem equation follows from, $$z^{13}=(p^2+n)^{15}-(p(p^2+n)^7)^2-((p^2+n)^2)^7-(p^2+n)^{14}$$ where $z$ is the $c$ of the original.
I’ve tried numerical calculations, and a little algebra, without progress.
Update 6 March 2017
I hope nobody objects, but I’ve now expanded this question from positive integer solutions to integer solutions. Although positive integer solutions (or proof there are none) are my main target, any results are welcome.
I can only see two trivial results, $$(n,p,z)=(2,a,0)$$ $$(n,p,z)=(-b^2,b,0)$$