Let p(1) < p(2) < p(3) < p(4) < p(5) < p(6) < p(7) be those primes. p(2)^8-p(1)^8=[p(2)-p(1)][p(2)+p(1)][p(2)^2+p(1)^2][p(2)^4+p(2)^4] p(i) > p(2) > p(2)-p(1) for every i, 2<i<8. Analogously, since all p(i) are odd and so are p(1) and p(2), all the p(i) dividing [p(1)^k+p(2)^k] must divide [p(1)^k+p(2)^k]/2 < p(2)^k < p(i1)p(i2)...p(ik), which means that at most k-1 of them divide [p(1)^k+p(2)^k] for each k. As a conclusion, p(i) divides p(2)^8-p(1)^8 for at most (1-1)+(2-1)+(4-1)=4<5 different values of i, and such a set cannot exist.

nice solution, Mimoide , but next time please use Latex.(just put dollar signs around the mathematical formulas and that would make your solutions easier to read)

My solution: $p_{1}\leq p_{2}\leq \dots\leq p_{7}$ then $\prod_{k=3}^{7}{p_{k}}|(p_{1}-p_{2})(p_{1}+p_{2})(p_{1}^{2}+p_{2}^{2})(p_{1}^{4}+p_{2}^{4})$ $p_{k}$ is odd then $\prod_{k=3}^{7}{p_{k}}|\frac{p_{1}-p_{2}}{2}*\frac{p_{1}+p_{2}}{2}*\frac{p_{1}^{2}+p_{2}^{2}}{2}*\frac{p_{1}^{4}+p_{2}^{4}}{2}$ $p_{3}>p_{1}-p_{2}$ $p_{3}>\frac{p_{1}+p_{2}}{2}$ so $\prod_{k=3}^{7}{p_{k}}|\frac{p_{1}^{2}+p_{2}^{2}}{2}*\frac{p_{1}^{4}+p_{2}^{4}}{2}$ but $\prod_{k=3}^{7}{p_{k}}>\frac{p_{1}^{2}+p_{2}^{2}}{2}*\frac{p_{1}^{4}+p_{2}^{4}}{2}$ a contradution.

let $p_{1}<p_{2}<...<p_{7}$ we have $p_{1}^8-p_{2}^8$=$(p_{1}^4+p_{2}^4)*(p_{1}^2+p_{2}^2)*(p_{1}+p_{2})*(p_{2}-p_{1})$ now , $p_{1}+p_{2}$ and $p_{2}-p_{1}$ are even so if they are divisible by $p_{k} , k=3,7$ they must be divisible by $2*p_{k}$ , but since they $p_{2}-p_{1} <p_{2}+p_{1}<2*p_{k}$ this is impossible so $\prod_{k=3}^{7}{p_{k}}|(p_{1}^{2}+p_{2}^{2})*(p_{1}^{4}+p_{2}^{4})$ but $p_{i}*p_{j}>2*p_{2}^2>p_{1}^2+p_{2}^2 , i,j=3,7$ so $p_{1}^2+p_{2}^2$ can be divisible by only one of the remainig primes thus , $p_{1}^4+p_{2}^4$ must be divisible by at least 4 of the reamining primes which is impossible

It is not necessary that the primes be odd. Suppose that $2 = p_1 < p_2 < \cdots < p_7$ as others have done. Then, as others have done \[ p_3 p_4 \cdots p_7 \mid (p_2^4 + 16)(p_2^2 + 4)(p_2 + 2) \]From here we split into two cases. Case 1: Suppose that $p_3 \neq p_2 + 2$. Then \[ p_3p_4p_5p_6p_7 \mid (p_2^4+16)(p_2^2+4) .\]However, $p_2^2 + 4$ can only have one divisors among $p_3, \dots, p_7$ since \[p_ip_j > (p_2 + 2)^2 > p_2^2 + 4. \]Likewise $p_2^4 + 16$ can only have three prime divisors among the $p_i$, so this case is finished. Case 2: SUppose that $p_3 = p_2 + 2$. In this case we obtain that \[ p_4 p_5 p_6 p_7 \mid (p_2^4+16)(p_2^2 + 4) . \]Additionally, since $p_4 \neq p_3 + 2$ because there are no triplet primes (the only exception is if $(p_2,p_3,p_4)=(3,5,7)$ but $7 \nmid 5^8 - 3^8$), we also obtain $p_4 p_5 p_6 p_7 \mid (p_3^4 + 16)(p_3^2 + 4)$. Then \begin{align*} p_4 p_5 p_6 p_7 &\mid (p_2^4+16)(p_2^2 + 4) - (p_3^4 + 16)(p_3^2 + 4) \\ &= p_2^6 - p_3^6 + 4p_2^4 - 4p_3^4 + 16p_2^2-16p_3^2 \\ &= (p_2^2 - p_3^2)(16 + 4p_2^2 + 4p_3^2 + p_2^2p_3^2 + p_2^4 + p_3^4) \end{align*}None of the primes can divide $p_2^2 - p_3^2$ for size reasons, so they must divide the second factor. Taken modulo 3, we can see that that term is divisible by 3, so in fact we ahve \[ p_4 p_5 p_6 p_7 \mid \frac{16 + 4p_2^2 + 4p_3^2 + p_2^2p_3^2 + p_2^4 + p_3^4}{3} < \frac{16 + 8p_3^2}{3} + p_3^4 < (2+p_3)^4\]which is a contradiction.