Show that the following system of congruences
\begin{align} \begin{cases} 3 x^4 - 7 x^2 y^2 - 7 x^2 z^2 - 35 y^2 z^2 \equiv 0 \pmod{p} \\ 3 y^4 - 7 x^2 y^2 - 7 y^2 z^2 - 35 x^2 z^2 \equiv 0 \pmod{p} \\ 3 z^4 - 7 y^2 z^2 - 7 x^2 z^2 - 35 x^2 y^2 \equiv 0 \pmod{p} \end{cases} \end{align}
has only the trivial solution when $p$ is a prime distinct from $2, 3, 5,$ and $23$.
Note: To obtain the restriction $p \neq 2, 23$, suppose that the system has a solution in which $x = y = z$. Then
\begin{align} 3 x^4 - 7 x^2 y^2 - 7 x^2 z^2 - 35 y^2 z^2 \equiv -46 x^4 \equiv 0 \pmod{p} \end{align}
has nontrivial solutions only when $p = 2, 23$.
I suspect that the symmetry in this system will help in showing that it has only the trivial solution at the other primes.
The symmetries are, indeed, the key. I let the variables range over the field $\Bbb{Z}_p$ to avoid having to write $\pmod p$ all the time.
Assume first that the squares $x^2,y^2,z^2$ are all pairwise non-congruent modulo $p$. Then subtracting your second equation from the first gives $$ 0=3x^4-3y^4+28x^2z^2-28y^2z^2=(x^2-y^2)(3x^2+3y^2+28z^2). $$ Given our assumption we can cancel the factor $x^2-y^2$ and end up with the congruence $$ 3x^2+3y^2+28z^2=0.\qquad(*) $$ Similarly operations with other pairs lead to the congruences, where the variables $x,y,z$ switch roles.
We thus arrive at a system of three congruences. These are "linear" in the unknowns $u=x^2$, $v=y^2$ and $w=z^2$. Let us denote the $3\times3$ matrix $$ A=\left(\begin{array}{ccc}28&3&3\\3&28&3\\3&3&28\end{array}\right). $$ We have $$ \det A=21250=2\cdot5^4\cdot17, $$ so the system $$ A\left(\begin{array}{c}u\\v\\w\end{array}\right)=\left(\begin{array}{c}0\\0\\0\end{array}\right) $$ has no non-trivial solutions unless $p\in\{2,5,17\}$. Two first of these possibilities were excluded at the starting gate. If $p=17$, then $A$ has rank $2$, so the solution space is 1-dimensional and clearly consists of the vectors $u=v=w$. But it is easy to check that the original system as no solutions with $x^2=y^2=z^2$ when $p=17$ (you already did that).
The remaining possibility is that some pair of squares of the variables are equal. By symmetry it suffices to handle the case $x^2=y^2$. In that case the original system becomes the pair $$ \begin{cases} -4 x^4 - 42 x^2 z^2 =0 \\ 3 z^4 - 14 x^2 z^2 - 35 x^4 =0 \end{cases} $$ If $x^2=0$ then the latter congruence implies $z=0$ also. Otherwise we can solve from the first congruence that $2x^2=-21z^2$. Multiplying the second congruence by four and plugging this in yields $14835z^4=0$. The original system has no non-trivial solutions with $x^2=y^2$ when $z=0$ so we are left with the case $$ 14835=3\cdot5\cdot7\cdot137=0. $$ In the case $p=7$ the original system immediately tells that only the trivial solution is there. The case $p=137$ remains. But if $p=137$ then $$ 2x^2\equiv -21 z^2\equiv116z^2\implies x^2\equiv 58z^2. $$ This has non-trivial solutions only if $58$ is a quadratic residue modulo $137$. It is easy to verify that this is not the case. Clearly $58=2\cdot29$. Because $137\equiv1\pmod 8$ we know that $2$ is a quadratic residue modulo $137$. With $29$ we use the law of quadratic reciprocity $$ \left(\frac{29}{137}\right)=\left(\frac{137}{29}\right)=\left(\frac{21}{29}\right). $$ As $29\equiv1\pmod7$ another round of reciprocity gives that $\left(\dfrac 7{29}\right)=1$. Similarly because $29\equiv2\pmod3$ we see that $\left(\dfrac 3{29}\right)=-1.$ The claim follows from this.