Note that: \begin{align*} x^5+x^4+1=x^5+x^4+x^3+1-x^3=\\ x^3(x^2+x+1)+(1-x)(1+x+x^2) =\\ =(x^3-x+1)(x^2+x+1) \end{align*}Consider the following cases: $x \equiv 1 \pmod 3$, then $x^2+x+1 \equiv 1+1+1 \equiv 0 \pmod 3$ and $x^3-x+1 \equiv 1-1+1 \equiv 1 \pmod 3$ $x \equiv 2 \pmod 3$, then $x^2+x+1 \equiv 2^2+2+1 \equiv 1 \pmod 3$ and $x^3-x+1 \equiv 2^3-2+1 \equiv 1 \pmod 3$ $x \equiv 0 \pmod 3$, then $x^2+x+1 \equiv 0^2+0+1 \equiv 1 \pmod 3$ and $x^3-x+1 \equiv 0-0+1 \equiv 1 \pmod 3$ This means that $3$ does not divide $x^3-x+1$, and $x^2+x+1$ is divisible by $3$ if and only if $x \equiv 1 \pmod 3$. Suppose that $x=3k+1$, then: $$ (3k+1)^2+(3k+1)+1 =9k^2+6k+1+3k+1+1=9k^2+9k+3 $$This means $x^2+x+1$ is not divisible by $9$. Therefore, $y=0$ or $y=1$. Now suppose that $d=\gcd(x^3-x+1,x^2+x+1)$. Then: $$ d \mid x(x^2+x+1) -(x^3-x+1)=x^2+2x-1 $$Observe that in that case: $$ d \mid x^2+2x-1 -x^2-x-1 =x-2 $$This means that $x \equiv 2 \pmod d \implies 0 \equiv 2^2+2+1 \equiv 7 \pmod d$. We conclude that $d=7$ or $d=1$. This means that one of the brackets is divisible by at most $7$ (but not by $49$). This gives us the following possibilities: $x^3-x+1 =1 $, then $x=0$ or $x=1$. In the former case, we get that $y=z=0$, whereas in the latter case $y=1$, $z=0$. $x^3-x+1 =7 \implies x^3-x-6=0$. The only integer root is $x=2$, which gives $y=0, z=2$. $x^2+x+1=1$, then $x=0$. This case has been considered before. $x^2+x+1=3$, then $x=1$. This case has been considered before. $x^2+x+1=7$, then $x=2$. This case has been considered before. $x^2+x+1=21$, then $x=4$. On the other hand, then $x^3-x+1 =61$ - not divisible by either $3$ or $7$. Thus, there are only $3$ possible triplets $(x,y,z)$, which are $(0,0,0)$, $(1,1,0)$ and $(2,0,2)$.