The formidable game of prime

Question

Does there exist any set of solution for equation $x^n*(y-u)=y^r-u$, where $x, y,r$ are prime numbers, $(x+1)$ and $(y-u)$ are powers of $2$, $u\in\{1,-1\}$ and $n$ is an integer except $(x,y,r,n,u)=(7,3,3,1,-1), (31,5,3,1,1)$?

Answer 1

One further solution is $(x,y,r,n,u)=(3,2,2,1,1)$. Other solutions may or may not exist, but note that you can almost always cancel the term $y-u$ from both sides by factoring the RHS of your equation. The one exception to this is when $r=2,u=-1$ in which case the RHS is $y^2+1$ and therefore the LHS becomes $x^n*(y+1)$. However, in this scenario there are no solutions at all for $x,y,n$ since the LHS will have a minimum $2^k$ factor of $4$ while the RHS will be of the form $2(2q+1)$.

One further group of solutions appears from $x$ in the set of Mersenne primes, $n=1,y=2,u=1$, then $r$ is the prime such that $x=M_r=2^r-1$, the Mersenne number at index $r$. The definition of $x$ appears to be the same as that of the Mersenne numbers.