Hint : $p>5$ so $p$ is not divisible by $3$, hence either $p-1$ or $p+1$ is divisible by $3$. So $3$ divides $p^4-1$. So $3$ divides it and $8$ divides it. Added: Notice that $p-1$ and $p+1$ are consecutive multiples of $2$, so one them is divisible by $4$. As Batimovski suggested, you can use Fermat's little theorem to conclude that $p^{4} \equiv 1$ mod $5$. So, we got so far $(p-1)(p+1)$ is divisible by $8$ and $3$, and $p^{2}+1$ is even, so $p^{4}-1$ divisible by $16$, and then by $3$, so by $16 \times 3 \times 5=240$ as they are pairwise comprime, the largest number on your list!