$24ml+1=k^2$ 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.
Under $100$ there are $84$ $m$-values yield no integer integers $(k)$ for $l=1$.
To find them is simple. $\quad k\not\in\mathbb{N} \text { for }k=\sqrt{24 m + 1)}$.
For example, for any non-zero integer $(m)$,
$$\text{If }\space \sqrt{24 m + 1)}\ne
\big\lfloor \sqrt{24 m + 1)}\space\big\rfloor 
\text{, then print } m \text{, endif }\tag{1}$$
This is the same as testing $\quad\sqrt{24 m \times(1) + 1)},\quad $
i.e. $l=1.\quad$ To find non-solution m-values for any other l-values, you must test the full $\quad 24m\times(l)+1, l\le m.\quad$ For examples,
\begin{align*}
\text{If }l=2,\quad&amp; m\in\{ \emptyset \}\\
\text{If }l=3,\quad&amp; m\in\{ 1,2,3\}\\
\text{If }l\in\{4,5,6,7,8\}\quad&amp; m\in\{ \emptyset \}\\
\text{If }l=9,\quad&amp; m\in\{ 1,2,3,4,5,6,7,8,9\}\\
\text{If }l\in\{10,11,12,\cdots, 23\}\quad&amp; m\in\{ \emptyset \}\\
\mathbf{\vdots}
\end{align*}
Your derived list is correct but I cannot offer a better formula for finding these m-values.
$\textbf{Update}\\$
We can solve for $l$ and see which k-values yield an integer for a given m-value.
$$l = \frac{k^2 - 1}{24 m} \quad\text{and we know }m\ne0\tag{2}$$
We can find $k^2$ by taking multiples of $24m$ and adding 1 to see if any $\sqrt{k^2}$ yields an integer. Having found $k$ we plug it back into equation $(2)$ and find $l$. We have a valid $m$ if and only if the smallest $l$ generated is greater than $m$. The test values for $m$ can quickly be obtained from equation $(1)$ above.
Under $100$ there are $84$ $m$-values yield no integer integers $(k)$ for $l=1$.
To find them is simple. $\quad k\not\in\mathbb{N} \text { for }k=\sqrt{24 m + 1)}$.
For example, for any non-zero integer $(m)$,
$$\text{If }\space \sqrt{24 m + 1)}\ne
\big\lfloor \sqrt{24 m + 1)}\space\big\rfloor 
\text{, then print } m \text{, endif }\tag{1}$$
This is the same as testing $\quad\sqrt{24 m \times(1) + 1)},\quad $
i.e. $l=1.\quad$ To find non-solution m-values for any other l-values, you must test the full $\quad 24m\times(l)+1, l\le m.\quad$ For examples,
\begin{align*}
\text{If }l=2,\quad&amp; m\in\{ \emptyset \}\\
\text{If }l=3,\quad&amp; m\in\{ 1,2,3\}\\
\text{If }l\in\{4,5,6,7,8\}\quad&amp; m\in\{ \emptyset \}\\
\text{If }l=9,\quad&amp; m\in\{ 1,2,3,4,5,6,7,8,9\}\\
\text{If }l\in\{10,11,12,\cdots, 23\}\quad&amp; m\in\{ \emptyset \}\\
\mathbf{\vdots}
\end{align*}
Your derived list is correct but I cannot offer a better formula for finding these m-values.
$\textbf{Update}\\$
We can solve for $l$ and see which k-values yield an integer for a given m-value.
$$l = \frac{k^2 - 1}{24 m} \quad\text{and we know }m\ne0\tag{2}$$
We can find $k^2$ by taking multiples of $24m$ and adding 1 to see if any $\sqrt{k^2}$ yields an integer. Having found $k$ we plug it back into equation $(2)$ and find $l$. We have a valid $m$ if and only if the smallest $l$ generated is greater than $m$. The test values for $m$ can quickly be obtained from equation $(1)$ above.