Cunning Repercussion

First note the expression $x^{5}y^{4}z^{3}+x^{4}y^{3}z^{2}+x^{3}y^{2}z+x^{2}y+x$ can be factored to $x(x^{4}y^{4}z^{3}+x^{3}y^{3}z^{2}+x^{2}y^{2}z$ +xy+1)) meaning $7907799$ must be divisible by $x$ and from the second equation we have that $x<=19$ . Checking $7907799$ for divisibility by primes under $19$ we find it only divisible by $3$ so $x=3$

Dividing the first equation by $x$ and subtracting $1$ yields:

$3^{4}y^{4}z^{3}+3^{3}y^{3}z^{2}+3^{2}y^{2}z+3y=2635932$

Divide by 3 again and factor out a $y$ :

$y(27y^{3}z^{3}+9y^{2}z^{2}+3yz+1)=878644$

Subsituting $x=3$ into the second equality we know $3y+4z=32$ giving us three potential options: $(y,z)=(0,8),(4,5),(8,2)$

But $y|878644$ so $y$ is certainly not $0$ and must be $4$ or $8$ , checking for divisibility by $8$ $878644=644=44=4 \mod 8$

We now have $x=3,y=4$ which implies $z=5$

The answer is then $x+y+z=3+4+5=\boxed{12}$