SFTSOC, they do. Note $a \neq d$. Since $a$ is bounded by $d^{2024}$, there must be infinitely many $d$. Consider a prime $q$, such that $q \nmid abcd$ and $q < abcd$. Observe that $abcd \leq d^{7073}$, and $d! > d^{7073} > q - 1$ for sufficiently large $d$, so $q - 1 \mid d! \implies d^{d!} \equiv 1\mod{q}$ by Fermat's little theorem. Repeat this with $a, b, c, d$ and we get $a^{a!}, b^{b!}, c^{c!}, d^{d!} \equiv 1\mod{q}$, so $q \mid a^{a!} + b^{b!} - c^{c!} - d^{d!}$. Consider another prime $r$ such that $r \nmid abcd$, and $r < abcd$, and we see that $r \mid a^{a!} + b^{b!} - c^{c!} - d^{d!}$, so our expression is not prime, a contradiction. Hence, we cannot have infinitely many quadruples