Keith Backman

@user1606702 Suppose $k$ is an integer. So what? Can it be any integer? Or can it be only certain integers? If it can be only one integer, then it is a constant. This is your question. It's not up to us to guess what you mean by "Suppose $k$ is an integer."

It is unclear what the question is. Do you ask if for any specific value of $k$ there are only finitely many $m,n$ that yield a solution, or do you ask if there are only finitely many $k$ for which a solution exists? Plainly, there are infinitely many $k$ for which at least one solution exists. Set $m=1$ and we have $k=n^2+1$ and there is a solution to that equation for every value of $n$.

It is conjectured but not proven that there are infinitely many Sophie Germain primes, i.e. primes $p$ such that $2p+1$ is also prime (called a safe prime).

@km2233 Oops! Comment deleted. Thanks for keeping me honest.

Either I misunderstand your proposition, or it is wrong. Take $N=22$. Then $P=2, Q=11$. But $\sqrt{22}-2>2$ and $11-\sqrt{22}>4$, so neither of your assertions are true, at least in this case.

HINT: $3^n \equiv (-1)^n \bmod 4$. Note that $(-1)$ raised to any odd power is always $(-1)$

Everywhere you use $\frac{m!}{m^2}$ where $m$ might refer to $t$, or $t+2$, or $t+k$ you could replace it with$\frac{ (m−1)!}{m}$. You are simply cancelling a factor of $m$ which is necessarily present in each such term, and it makes it more obvious why the term is not an integer when $m$ is prime, and is an integer when $m$ is composite (other than the special case of $4$).

My bad; I misunderstood the limitation.