Solving the equation $yz=y^z-2$ 
Question: Let $x,y$ and $z$ be positive integers. How can we determine the positive integer solutions to the equation $$yz=y^z-2$$

The motivation here is straightforward. Surely $6=2\times 3$ and also $8=2^3$ and $9=3^2$. I wanted to "generalize" those observations by writing 
\begin{align}
x&=yz\\
x+2&=y^z\\
x+3&=z^y
\end{align}
Trivially we can get to $y^z-2=x$ and after substitution we have the equation $$yz=y^z-2$$
WolframAlpha quickly spits out $y=2$ and $z=3$ as the only positive integer solutions. It also gives a "solution for the variable z" as 
$$z=-{yW\left(-y^{-1-{2\above 1.5pt y}} \right)-2\log{y}\above 1.5pt y\log{y}} $$
where $W(a)$ is the Lambert W function. Following the Wiki link leads me to think a possible start is to consider $$-t=x+{2\above 1.5pt y} $$ and we can now transform our equation into $$ty^t=-{1\above 1.5pty}y^{-{2\above 1.5pt y }}$$ which apparently yields the solution for $z$ written above. But I am still not clear how WolframAlpha explicitly determined that $z=3$. 
 A: Lemma(I): Let $3 \leq y$, then we have:
$$2y < y^2-2.$$
Proof: $3 \leq y$, so we must have: 
$$2 \leq (y-1) 
\Longrightarrow 
3 < 4 \leq (y-1)^2 
\Longrightarrow 
3 < y^2-2y+1 
\Longrightarrow 
2y < y^2-2 . 
$$

Lemma(II): 
Let $3 \leq y$ be any fixed arbitrary integer 
and let $2 \leq z$, then we have: 
$$  \color{Red}{yz < y^z-2}  \ \ \ \ \ \ \ \ \ \  \text{(II)}   .   $$
Proof: 
Let $y$ be an  arbitrary (but fixed) integer 
greater or equal than $3$. 
We will prove the lemma by induction on $z$.
$z=2$; which is done by Lemma(I). 
Now suppose that the Lemma(II) is true for $z=k$; 
then we will show it for  $z=k+1$. 
Only notice that: 
$$ \color{Blue}{y} < zy+2 < y^z < (y-1)y^z = 
\color{Blue}{y^{z+1}-y^z}
\ 
; $$
adding this last inequality by inequlaity (II) we get: 
$$ y(z+1)= \color{red}{yz}+ \color{Blue}y <  
\color{Red}{y^z-2} +   \color{Blue}{y^{z+1}-y^z} = 
y^{z+1}-2  .$$

Lemma(III): 
Let $y=2$ 
and let $4 \leq z$, then we have: 
$$  \color{Red}{2z < 2^z-2}  \ \ \ \ \ \ \ \ \ \  \text{(III)}   .   $$
Proof:
We will prove the lemma by induction on $z$.
$z=4$; is trivial. 
Now suppose that the Lemma(III) is true for $z=k$; 
then we will show it for  $z=k+1$. 
Only notice that: 
$$ \color{Blue}{2} < 2z+2 < 2^z = 
\color{Blue}{2^{z+1}-2^z}
\ 
; $$
adding this last inequality by inequlaity (III) we get: 
$$ 2(z+1)= \color{red}{2z}+ \color{Blue}2 <  
\color{Red}{2^z-2} +   \color{Blue}{2^{z+1}-2^z} = 
2^{z+1}-2  .$$




First Case: Let $3 \leq y$: 


*

*$2 \leq z$; which is impossible by Lemma(II).

*If $z=1$; then we have $y.1=y^1-2$, 
which is obviously impossible again!


Second Case: Let $y=2$: 


*

*$4 \leq z$; which is impossible by Lemma(III).

*$z=3$; which gives the solution $\color{Green}{(y,z)=(2,3)}.$

*$z=2$; which does not give any solution!  

*$z=1$; which does not give any solution again! 
Third case: Let $y=1$: 


*

*in this case we have: $1.z=1^z-2$; 
which does not have a solution in $\mathbb{N}$.

A: Let $y$ and $z$ be positive integers such that $yz=y^z-2$. Then also
$$y(y^{z-1}-z)=2,$$
which shows that $y$ divides $2$, so either $y=1$ or $y=2$. Correspondingly we get
$$1^{z-1}-z=2\qquad\text{ and }\qquad 2^{z-1}-z=1,$$
and the former clearly has no integral solutions. The latter is equivalent to
$$2^{z-1}=z+1,$$
which is easily seen to have $z=3$ as the only integral solution.
