$$P(m,n)\Rightarrow f (mf (n + 1)) = f (m + 1) f (n) + f (f (n)) + 1 $$$$P(0,n)\Rightarrow f(0)=f(1)f(n)+f(f(n))+1$$$$P(1,n)\Rightarrow f(f(n+1))=f(2)f(n)+f(f(n))+1$$But $f(f(n))=f(0)-f(1)f(n)-1$, now: $$f(f(n+1))=f(0)-f(1)f(n+1)-1$$Then: $$f(0)-f(1)f(n+1)-1=f(2)f(n)+f(0)-f(1)f(n)-1+1$$$$-f(1)f(n+1)=f(2)f(n)-f(1)f(n)+1$$Now, in that equation take $n=1$, then: $$-f(1)f(2)=f(2)f(1)-f(1)^2+1$$Let $a=f(1)$ and $b=f(2)$ then: $$-ab=ab-a^2+1\Rightarrow 2b=\frac{a^2-1}{a}=a-\frac{1}{a}$$Then $b=0$ and $a=\pm 1$. Case 1, $a=1$ $$-f(n+1)=-f(n)+1$$With a easy ''doble'' induction we obtain $f(n)=2-n$ for all $n\in \mathbb{Z}$. Case 2, $a=-1$ $$f(n+1)=f(n)+1$$And analogous to case $1$, with a easy induction we have $f(n)=n-2$ But none of them satisfy the functional equation, so there is no function that satisfies that

Can someone help me to finish the solution? How can I find a contradiction for $y<1$? $1)$ $P(0,n) \implies f(0)=f(1)f(n)+f(f(n))+1$ $2)$ $P(1,n) \implies f(f(n+1))=f(2)f(n)+f(f(n))+1$ $1,2$ together implies $f(1)f(n+1)=(f(1)-f(2))f(n)-1$ $i) f(1)=0$ $f(f(n))=f(0)-1\implies f(2)f(n)=-1 \implies f(2) \neq 0 \implies f(n)=c \implies$ contradiction. $ii) f(1)\neq 0$ $f(n+1)=yf(n)+z$.If $y=0,f(n)$ is constant which is not a solution. If $y=1$, we easily have $f(n)$ linear which is not a solution. Otherwise there exists $t$ such that $t(y-1)=z$. We have $f(n+1)+t=y(f(n)+t) \implies g(n+1)=yg(n) \implies g(n)=g(0)y^n \implies f(n)=cy^n-t$. Plug this in $1 \implies c=(cy-t)(cy^n-t)+cy^{cy^n-t}$. If $y>1$ and $n$ is big enough we have a contradiction,if $y<1$ ???. Hence no solution.

$P(0,n)\Rightarrow f(f(n))=f(0)-f(1)f(n)-1$ $P(1,n)\Rightarrow f(f(n+1))=f(2)f(n)+f(f(n))+1\Rightarrow f(1)f(n+1)=(f(1)-f(2))f(n)-1~(\star)$ Consider three trivial cases: If $f(1)=f(2)=0$, then $P(1,1)\Rightarrow f(0)=1$ and $P(0,2)\Rightarrow f(0)=0$, contradiction. If $f(1)=0$ and $f(2)\ne0$, then $(\star)$ implies that $f$ is constant, so no solutions. If $f(1)=f(2)\ne0$, then $(\star)$ implies that $f$ is constant. The only alternative to these three is that $0\ne f(1)\ne f(2)\ne0$, so that $f$ is linear by $(\star)$. Then $\boxed{\text{no solutions}}$ exist.

$$f(n)(f(1)-f(2))=f(1)f(n+1)+1$$Remaining easy