$24ml+1=k^2$ has no solution for all $l=1 \dots m$ Investigating solutions of $$24ml+1=k^2$$ for $l=1\dots m$
The question is to find the $m$-s for which the above equation has no solution for all $l=1..m$-s.
The first few $m$-s are:
$$3, 9, 24, 27, 81, 192, 243, 432, 729$$
Actually have found that the $m$ should be of form $2^a3^b$. Seems hang on the simple task, of finding general formula for this.

One interesting addition. Just tested the cases when the $$12ml+1=k^2$$ has no solution for all $$l=1..m$$ Seems it has no solution iff $m=3^a$ But has no proof yet.

Here are a couple of results that can be proved:
There is no solution if $m=3^a, a>0.$
For $24ml=(k-1)(k+1)$ we must have $k=6x\pm 1$. Then$$ml=\frac {x(3x\pm 1)}{2}.$$
Now if $m=3^a, a>0$, then $m$ has to be a factor of $x$. But then $l\ge \frac {3x- 1}{2}>m,$ a contradiction.
There is a solution if $m$ is not divisible by $3$.
Let $2m=3l\pm 1$. Then  $l\le m$ and
$$24lm+1=12l(3l\pm 1)+1=(6l\pm 1)^2.$$
