Let $3^a5^b-2024=x^2$. $3^a \equiv {1,3}$ and $5^b \equiv {1,5} \pmod8$. Because $x^2 \equiv {0,1}\pmod8$ the only valid case is $3^a \equiv 1$ and $5^b \equiv 1$ which means $a,b$ are even. Let $a=2a_1$ and $b=2b_1$. Then we have $2024 = (3^{a_1}5^{b_1}-x).(3^{a_1}5^{b_1}+x)$. From simple case bash to divisor of 2024, we can get $(a,b)=(4,2)$ is the only solution. $\blacksquare$

sevket12 wrote: Let $3^a5^b-2024=x^2$. $3^a \equiv {1,3}$ and $5^b \equiv {1,5} \pmod8$. Because $x^2 \equiv {0,1}\pmod8$ the only valid case is $3^a \equiv 1$ and $5^b \equiv 1$ which means $a,b$ are even. Let $a=2a_1$ and $b=2b_1$. Then we have $2024 = (3^{a_1}5^{b_1}-x).(3^{a_1}5^{b_1}+x)$. From simple case bash to divisor of 2024, we can get $(a,b)=(4,2)$ is the only solution. $\blacksquare$ amazing solution

sevket12 wrote: Let $3^a5^b-2024=x^2$. $3^a \equiv {1,3}$ and $5^b \equiv {1,5} \pmod8$. Because $x^2 \equiv {0,1}\pmod8$ the only valid case is $3^a \equiv 1$ and $5^b \equiv 1$ which means $a,b$ are even. Let $a=2a_1$ and $b=2b_1$. Then we have $2024 = (3^{a_1}5^{b_1}-x).(3^{a_1}5^{b_1}+x)$. From simple case bash to divisor of 2024, we can get $(a,b)=(4,2)$ is the only solution. $\blacksquare$ The case bash would lighten significantly considering the fact that $$(3^{a_1}5^{b_1}-x,3^{a_1}5^{b_1}+x)\mid 2\cdot3^{a_1}5^{b_1}.$$Both $3$ and $5$ does not divide $2024$ so the GCD can only be equal to $1$ or $2$. Case bash after this observation directly translates to my in-contest solution.

The same as #2, posting for storage.