$24ml+1=k^2$ 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.

The OP also conjectures that $12ml+1=k^2$ has no solutions for $1\le l\le m$ if and only if $ m=3^a$. This is true and can be proved as follows.
Only if
For $12ml=(k-1)(k+1)$ we must have $k=6x\pm 1$. Then $ml=x(3x\pm 1).$
Let $m=3^aM$, where $M>1$ is coprime to $3$. Let $x=3^aX$ and then $$Ml=X(3^{a+1}X\pm 1).$$
We can choose $X$, $0< X <M$, such that $3^{a+1}X\equiv 1$ modulo $M$. Replacing $X$ by $M-X$ if necessary we can therefore choose $X$,  $0< X \le \frac{M}{2},$ such that $(3^{a+1}X\pm 1)=MY$ for some integer $Y$.
Then $l=XY$.
If $m<l$ then $$3^aM<\frac{M}{2}\frac{(3^{a+1}M\pm 2)}{2M}  \text{ i.e. }  3^a<\frac{3^{a+1}\pm \frac{2}{M}}{4}.  $$
Then $a=0$ and $M=2$. But then $l=1$ gives the solution $12ml+1=5^2$.
If
Let $m=3^a$ and $x=3^aX$,  then we must solve $l=X(3^{a+1}X\pm 1)>3^a=m,$ a contradiction.
