Vieta jumping. Solutions exists only for $z=7$ There are two series of solutions with first terms $1,2$ and $1,4$

$\wedge$ means 'and'. $\mathbb{N*}$ means $\{n | n\in \mathbb{Z} \wedge n>0\}$. $(*)$ stands for the equation $x^2+y^2+x+y+z=xyz+1$. Define $g(x,y):=\frac{x^2+y^2+x+y-1}{xy-1}$. WLOG assume $x\geq y$. $\textbf{Case 1.}$ $y=1$. $x^2+x+z+2=xz+1 \Rightarrow (x-1)z=x^2+x+1>0 \Rightarrow x>1$. $z=g(x,1)=\frac{x^2+x+1}{x-1}=x+2+\frac{3}{x-1} \in \mathbb{Z}$ $\Rightarrow x=2, 4$. $(x,y,z)=(2,1,7), (4,1,7)$. $\textbf{Case 2.}$ $x=y\geq 2$. $g(x,x)=\frac{2x^2+2x-1}{x^2-1}=2+\frac{2x+1}{x^2-1} \Rightarrow 2x+1 \geq x^2-1 \Rightarrow x=2$. But when $x=y=2$, $g(x,x)=2+\frac{5}{3} \notin \mathbb{N*}$, so no triplets satisfy (*) in this case. Thus we have $x>y$. $\textbf{Case 3.}$ $y=2$. $g(x,2)=\frac{x^2+x+5}{2x-1}$. $4g(x,2)=\frac{4x^2+4x+20}{2x-1}=2x+3+\frac{23}{2x-1}\in \mathbb{N*}$ $\Rightarrow (2x-1)|23 \Rightarrow x=12\Rightarrow (x,y,z)=(12,2,7)$. $\textbf{Case 4.}$ $y\geq 3 \wedge x=y+1$. $g(y+1,y)=\frac{2y^2+4y+1}{y^2+y-1}=2+\frac{2y+3}{y^2+y-1}\in \mathbb{N*}$ $\Rightarrow 2y+3\geq y^2+y-1$ which contradicts with $y\geq 3$. $\textbf{Case 5.}$ $y\geq 3 \wedge x\geq y+2$. Suppose $(x,y,z)$ satisfies $x^2+y^2+x+y+z=xyz+1 (*)$. $z=g(x,y)=\frac{x^2+y^2+x+y-1}{xy-1}\geq \frac{2xy+x+y-1}{xy-1} = 2+\frac{x+y+1}{xy-1} > 2 \Rightarrow z\geq 3$. Fix $y,z$, and then $a^2-(yz-1)a+y^2+y+z-1=0$ is a quadratic, with one root $x$ and so the other is $\frac{y^2+y+z-1}{x}=yz-x-1$. Because $y^2+y+z-1 > 0$, $yz-x-1 \in \mathbb{N*}$. \begin{align*} y'<y\quad \Leftrightarrow\quad &yz-1-x<y \\ \Leftrightarrow\quad &y\frac{x^2+y^2+x+y-1}{xy-1}<x+y+1 \\ \Leftrightarrow\quad &x^2y+y^3+xy+y^2-y<x^2y+xy^2+xy-x-y-1 \\ \Leftrightarrow\quad &y^3+y^2-y<xy^2-x-y-1 \\ \Leftarrow\quad &y^3+y^2-y<(y+2)(y^2-1)-y-1 \\ \Leftrightarrow\quad &y^2-y-3\geq 0\\ \Leftarrow\quad &y\geq 3. \end{align*} \begin{align*} x'-y'<x-y\quad\Leftrightarrow\quad &y-(yz-x-1)<x-y \\ \Leftrightarrow\quad &yz>2y+1 \\ \Leftarrow\quad &z\geq 3 \wedge y\geq 2 \end{align*} Define $f(x,y,z):=(y,yz-x-1,z)$. Then if $(x,y,z)$ satisfies (*), so does $f(x,y,z)$. Suppose $(x,y,z)$ satisfies (*), and then we can replace it with $f(x,y,z)$ finite times until it does not satisfy the condition $x-2\geq y\geq 3$, because each time $y$ and $x-y$ strictly decreases. At last we will certainly have $(x,y,z)=(2,1,7),(4,1,7)$ or $(12,2,7)$. Surprisingly $f(12,2,7)=(2,1,7)$. Thus for all $(x,y,z)$ satisfying (*), we can replace it with $f(x,y,z)$ finite times to become (2,1,7) or (4,1,7). (Thus $z=7$.) If $(x,y,z)\in \mathbb{N*}^3$, then $f^{-1}(x,y,z)=(xz-y-1,x,z)\in \mathbb{N*}^3$. Let $(x_0,y_0)=(2,1)$ or $(4,1)$. Let $(x_{n+1},y_{n+1})=(7x_n-y_n-1,x_n)$. Then the two sequences of triplets $(x_n,y_n,7)$ (with different $(x_0,y_0)$) contains "all" the triplets satisfying (*). Be careful! $(y_n,x_n,7)$ satisfies (*) as well, because WLOG at the beginning. We have $y_{n+1}=x_n$ and so $y_{n+2}=7y_{n+1}-y_n-1$. The only thing left to do is to solve two sequences.