Primes of the form $x^3+y^3+z^3 - 3xyz$ Do quadruplets $(x,y,z,p)$ of positive integers exist for which $p$ is a prime number and $$x^3+y^3+z^3 = 3xyz + p?$$ 
I've tried looking for solutions in mathematica for $x,y,z<1000$, without finding any. Unfortunately, looking at the equation $\mod 3$ or $\mod 9$ has yielded no result. Is there anyone who can help? Thanks.
EDIT: I just discovered a fatal flaw in my mathematica code. vrugtehagel rightfully pointed out that $8^3+8^3+7^3 - 3\cdot8\cdot8\cdot7 = 23$ is a valid solution to this problem.
 A: We use the identity
$$x^3 + y^3 + z^3 - 3xyz = (x+y+z)\left( \frac{ (x-y)^2 + (y-z)^2 + (z-x)^2 }{2}\right)$$
to try to find some solutions using linear and quadratic equations with the following system:
$$\begin{cases}x+y+z=1\\(x-y)^2+(y-z)^2+(z-x)^2=2p\end{cases}$$
Putting   $z=1-x-y$ one gets 
$$(x-y)^2 + (y-z)^2 + (z-x)^2=2(x^2+y^2+z^2)-2(xy+xz+yz)=2p\Rightarrow 3(x^2+y^2+xy-x-y)+1=p$$
We look for primes in the arithmetic progression $\{3n+1\}_{n\in \mathbb N}$ so we get the equation $$x^2+y^2+xy-x-y=n$$ and we look for integer points in ellipses.
We look at the six first  primes $p=7,13,19,31,37,43$ of the form $3n+1$ corresponding to $n=2,4,6,10,12,14$ respectively. 
$$x^2+y^2+xy-x-y=2\qquad (*)$$
Solutions $(x,y)=(0,-1),(2,-1),(2,0)$ yield  the solution $(x,y,z)=(2,-1,0)$ giving an example for
$$x^3+y^3+z^3-xyz=7$$
$$x^2+y^2+xy-x-y=4\qquad (**)$$
Solutions $(x,y)= (1,\pm2),(2,-2)$ yield the solutions $(x,y,z)=(1,\pm 2,\mp 2),(2,-2,1)$ giving examples for $$x^3+y^3+z^3-xyz=13$$
$$x^2+y^2+xy-x-y=6\qquad (***)$$  Solutions $(x,y)=(0,-2),(3,-2),(3,0)$ leads to examples for $p=19$.
Similarly $$x^2+y^2+xy-x-y=10, 12, 14\qquad (****)$$  admit respectively the solutions
$\{(-1,-2),(4,-2),(4,-1)\},\{(0,-3),(4,-3),(4,0)\}$  and $\{(2,-4),(3,-4),(3,2)$ which give solutions for 
$$x^3+y^3+z^3-xyz=31,37\space \text{and} \space 43$$
FINAL NOTE.- I wanted just to exhibit some examples of solutions like the case $p=23$ given by the O.P.(note that prime $23=3\cdot 7+2$ it is not of the form $3n+1$). However I add a question and a pertinent quote to finish.
►“It is very difficult to prove the non-existence of integer solutions of the general equation
$$ax^3+by^3+cz^3-dxyz=0”\space \text{(L.J.Mordell)}$$
►Is there always solution for $x^3+y^3+z^3-xyz=p$ when $p=3n+1$? (for the first 6 examples of such primes, I have found without difficulty solutions). 
A: $${x}^{3}-3\,y\,z\,x+{z}^{3}+{y}^{3}=\left( x+z+y\right) \,\left( {x}^{2}-z\,x-y\,x+{z}^{2}-y\,z+{y}^{2}\right) $$
$${x}^{2}-z\,x-y\,x+{z}^{2}-y\,z+{y}^{2}=1$$
if x,y,z - consecutive terms in http://oeis.org/A004526
then $p=3\,n+1,p=3\,n+2$
