Solving a Diophantine equation with power of $24$. We are going to prove that there is no integer solution for the Diophantine equation
$$\displaystyle 24^{a}=3 b^{2}+3 b \tag*{(*)} \\$$
$\textrm{Splitting }24=3\times 8 \text{ and factorising }3b^2+3b \textrm{ yields }$$ 3^{a-1} \cdot 8^{a}=b(b+1).$
$ (b, b+1)=(b,1)=1 \textrm{ and } 3^{a-1}<8^a \Rightarrow  b = 3 ^ { a - 1 } \textrm{ and }  b + 1 = 8 ^ { a } \Rightarrow 3^{a-1} +1=8^{a} \\  3^{a-1}=8^{a}-1  =7\left(8^{a-1}+8^{a-2}+\cdots+1\right)  \Rightarrow \quad 7 \mid 3^{a-1}, \text { which is a contradiction. } \tag*{} $
Hence we can conclude that the Diophantine equation (*) has no integer solution.
Your comments and alternate solutions are highly appreciated.
 A: Treat the equation as a quadratic equation in variable $b$. Thus: $b^2+b - 2^{3a}\cdot 3^{a-1} = 0$. Taking usual $\triangle = 1-4(1)(-2^{3a}\cdot 3^{a-1})=1+2^{3a+2}\cdot 3^{a-1}=k^2$, for some integer $k$. Hence: $(k-1)(k+1) = 2^{3a+2}\cdot 3^{a-1}$. Observe that $\text{gcd}(k-1,k+1) = 1$ or $2$. If $\text{gcd}(k-1,k+1) = 1$, then these are coprime integers. Thus: $k-1 = 2^{3a+2}, k+1 = 3^{a-1}$, or $k-1 = 1, k+1 = 2^{3a+2}\cdot 3^{a-1}$ or $k-1 = 3^{a-1}, k+1 = 2^{3a+2}$. From this, you can show either case yields no integer solutions in $k, a$. If $\text{gcd}(k-1,k+1) = 2$, then it must be the case that: $k-1 = 2^{3a+1}, k+1 = 2\cdot 3^{a-1}$ or $k-1 = 2\cdot 3^{a-1}, k+1 = 2^{3a+1}$. And again, either case yields no integer solutions in $a,k$. So overall, there is no integer solutions in $a,b$.
Note: We can demonstrate one example above that for the case $k-1 = 2^{3a+2}, k+1 = 3^{a-1}\implies 2k = 2^{3a+2}+3^{a-1}$. This has no solution since the left is even while the right is odd. The remaining cases are done similarly.
