I claim $(a,b,c)=(5,8,14)$ is the only solution. I first show $a\ge 6$. Suppose $a\ge 7$. Then, $b\ge a\ge 7$ yields $7\mid c^4+2024$. Since $2024\equiv 1\pmod{7}$, this forces $c^4\equiv -1\pmod{7}$, which together with $c^6\equiv 1\pmod{7}$ (Fermat) yields $c^2\equiv -1\pmod{7}$, not possible. Hence, $a\le 6$. Furthermore, it's also clear that $b!\ge 1012$ so $b\ge 7$. Next, using modulo 5, we find that $c^4+2024\in\{0,4\}\pmod{5}$, so $a\in\{4,5,6\}$. Now, let $a=6$. Then, $b!=c^4+1304$. It is evident that $16\mid b!$ and $c$ is even, but $v_2(1304)=3$, yielding a contradiction. Next, let $a=5$, yielding $b!=c^4+1904$. Notice that $7\mid b!$ and $7\mid 1904$, so $7\mid c$. As $v_7(1904)=1$, it follows that $v_7(b!)=1$, yielding $b\in\{7,\dots,13\}$. In this range, $b=8$ is the only solution, achieved by the triple $(a,b,c)=(5,8,14)$. Lastly, we study the case $a=4$, yielding $b!=c^4+2000$. If $b\ge 13$, then $c^4\equiv 2\pmod{13}$, yielding $c^{12}\equiv 8\pmod{13}$, not possible. So, $b\le 12$, which brings no solutions.

\[a!+b!=c^4+2024\]$16$ doesn't divide $RHS$ hence $a\leq 5$. $i)a=1\iff c^4+2023=b!\implies b\leq 2$ since $LHS\not \equiv 0(mod \ 3)\implies $ No solution. $ii)a=2\iff c^4+2022=b!\implies b\leq 3$ since $LHS\not \equiv 0(mod \ 4)\implies $ No solution. $iii)a=3\iff c^4+2018=b!\implies b\leq 3$ since $LHS\not \equiv 0(mod \ 4)\implies $ No solution. $iv)a=4\iff c^4+2000=b!\implies b\leq 10$ since $LHS\not \equiv 0(mod \ 11)\implies $ No solution. $v)a=5\iff c^4+1904=b!$ If $b\leq 7,$ we have no solution. $a=5,b=8,c=14$ is a solution. Suppose that $b\geq 9$. If $v_2(c)\geq 2,$ then $v_2(c^4+1904)=4\implies b\leq 7$ which gives a contradiction. Hence $c^4=16k^4$ \[16(k^4+119)=b!\]If $b\geq 10,$ then $v_2(b!)\geq 8\implies 16|k^4+119\implies 16|k^4+7$ which is impossible. $b=9$ gives no solution. Below, thanks.

bin_sherlo wrote: \[a!+b!=c^4+2024\]$16$ doesn't divide $RHS$ hence $a\leq 5$. $i)a=1\iff c^4+2023=b!\implies b\leq 2$ since $LHS\not \equiv 0(mod \ 3)\implies $ No solution. $ii)a=2\iff c^4+2022=b!\implies b\leq 3$ since $LHS\not \equiv 0(mod \ 4)\implies $ No solution. $iii)a=3\iff c^4+2018=b!\implies b\leq 3$ since $LHS\not \equiv 0(mod \ 4)\implies $ No solution. $iv)a=4\iff c^4+2000=b!\implies b\leq 6$ since $LHS\not \equiv 0(mod \ 7)\implies $ No solution. $v)a=5\iff c^4+1904=b!$ If $b\leq 7,$ we have no solution. $a=5,b=8,c=14$ is a solution. Suppose that $b\geq 9$. If $v_2(c)\geq 2,$ then $v_2(c^4+1904)=4\implies b\leq 7$ which gives a contradiction. Hence $c^4=16k^4$ \[16(k^4+119)=b!\]If $b\geq 10,$ then $v_2(b!)\geq 8\implies 16|k^4+119\implies 16|k^4+7$ which is impossible. $b=9$ gives no solution. I think for point 4 it is supposed to be mod 11 instead of mod 7

What a dizzying Diophantine. This was quite extreme. We claim that the only triple $(a,b,c)$ of solutions is $(5,8,14)$. It is easy to see that this triple indeed satisfies the given equation. Now we shall show that it is the only one. We start off by bounding $b$. Clearly, \[2024 < c^4 + 2024 = a! + b! \leq 2b!\]which implies $b! \geq 1012$, so $b \geq 7$. Now, if $c \equiv 1 \pmod{2}$, $c^4 + 2024 \equiv 1 + 2024 \equiv 1 \pmod{2}$. Further, $a!+b! \equiv a! \pmod{2}$ since $b>1$. Thus, $a! \equiv 1 \pmod{2}$ which requires $a=1$. Thus, \[b! + 1 \equiv c^4 + 2024 \implies b! \equiv 1 +2023 \equiv 8 \pmod{16}\]But, since $b\geq 7$, $16 \mid b!$ which is a clear contradiction. So, $c$ must be even. Then, $16 \mid b!$ as before and, \[8 \equiv 2024 \equiv c^4 + 2024 = a! + b! \equiv a! \pmod{16}\]Thus, since $16 \nmid a!$, $a! <6$. Further, we require $8 \mid a!$ which implies $a\geq 4$. Thus, $a=4$ or $a=5$. Now, remembering that $2024$ is divisible by $11$, we note that if $b \geq 11$, \[a! + b! \equiv a! \in \{2,-1\} \pmod{11}\]But, this means we need $c^4 \in \{2,-1\} \pmod{11}$ which is a clear contradiction since neither is a possible residue for a perfect fourth power$\pmod{11}$. Thus, we need $b<11$. We now are left with the dirty work. Case 1 : $a=4$. Then, the given equation rewrites to \[b! = c^4 + 2000\]But, clearly $10 \mid b$ since $b>5$. So, $10 \mid c^4$ which implies $10 \mid c$. But then, $1000 \mid c^4$ which implies $1000 \mid c^4 + 2000$. But, since $b<11$, $b!$ is not divisible by $1000$, so this case has no solutions. Case 2 : $a=5$. Then, the given equation rewrites to \[b! = c^4 + 1904\]Then, checking the only cases left , $b \in \{7,8,9,10\}$ we see that $b=8$ only gives a working solution, which is indeed the desired triple.

Suppose $a\geq 6$ - then $a!$ and $b!$ are divisible by $16$, while $2024$ is not divisible by $16$ but is by $8$, thus $c^4$ must be divisible by $8$ but not $16$, impossible. If $a=5$, then $b! = c^4 + 1904$ and here $b\geq 7$ and mod $7$ eliminates $b\geq 14$ since $1904 = 2^6 \cdot 7 \cdot 17$ is divisible by $7$, but not by $49$, while a check with $b=8,9,\ldots,13$ yields only $b=8$ with $c=14$. If $a=4$, then $b! = c^4 + 2000$ and here $b\geq 7$ and mod $5$ we have that $c$ is divisible by $5$, then mod $5^3$ yields $b\geq 15$, but $b\geq 20$ is eliminated by mod $5^4$, while $b=15,16,17,18,19$ can be verified directly (or eliminated by mod $11$ or mod $13$ if you really want to avoid calculations). If $a\leq 3$, then $b! \geq 2018$ requires $b\geq 7$, so mod $5$ requires $c$ to be not divisible by $5$ and hence $c^4 \equiv 1 \pmod 5$, yielding a contradiction for each of $a=1,2,3$.