When will this quadratic be a perfect square?

Question

What are all the positive integer solutions to the equation $m^2 + m - 1$, such that the output is a square number? I can tell that $m-1$ must be the negative or positive sum of odd numbers, so $m-1$ must be in the form $2mk + k^2$ where k is any integer. So we are left with $2m(k-1) + k^2 + 1 =0$, in which k can be any integer and m must be positive integer, but I can't seem to get anywhere with this unfortunately. Any help? Might it be useful that $|m-1|$ must be the difference of two squares?

Answer 1

We have $$m^2 < m^2 + m - 1 < m^2 + 2m + 1 = (m + 1)^2$$ for $m \ne 1$. This means that for $m \ne 1$ the number $m^2 + m - 1$ lies strictly between two consecutive squares and therefore cannot be a square itself.