If we add 2 to both sides, and factor, we obtain:
$$(t-2)(z-1)(xy+1)=14$$
Looking at the permutations of the factorization $14=1\cdot1\cdot14$, we find:
i) $$t-2=1\implies t=3$$
$$z-1=1\implies z=2$$
$$xy+1=14\implies xy=13\implies (x,y)=(1,13),\,(x,y)=(13,1)$$
ii) $$t-2=1\implies t=3$$
$$z-1=14\implies z=15$$
$$xy+1=1\implies xy=0$$ This case doesn't meet the requirements given.
The third permutation won't work for the same reason.
Looking at the permutations of the factorization $14=1\cdot2\cdot7$, (where the third factor isn't 1) we find:
i) $$t-2=1\implies t=3$$
$$z-1=2\implies z=3$$
$$xy+1=7\implies xy=6\implies (x,y)=(1,6),\,(6,1),\,(2,3),\,(3,2)$$
ii) $$t-2=1\implies t=3$$
$$z-1=7\implies z=8$$
$$xy+1=2\implies xy=1\implies (x,y)=(1,1)$$
iii) $$t-2=2\implies t=4$$
$$z-1=1\implies z=2$$
$$xy+1=7\implies xy=6\implies (x,y)=(1,6),\,(6,1),\,(2,3),\,(3,2)$$
iv) $$t-2=7\implies t=9$$
$$z-1=1\implies z=2$$
$$xy+1=2\implies xy=1\implies (x,y)=(1,1)$$
Thus, we find the following 12 solutions:
$$(t,x,y,z)=(3,1,13,2)$$
$$(t,x,y,z)=(3,13,1,2)$$
$$(t,x,y,z)=(3,1,6,3)$$
$$(t,x,y,z)=(3,6,1,3)$$
$$(t,x,y,z)=(3,2,3,3)$$
$$(t,x,y,z)=(3,3,2,3)$$
$$(t,x,y,z)=(3,1,1,8)$$
$$(t,x,y,z)=(4,1,6,2)$$
$$(t,x,y,z)=(4,6,1,2)$$
$$(t,x,y,z)=(4,2,3,2)$$
$$(t,x,y,z)=(4,3,2,2)$$
$$(t,x,y,z)=(9,1,1,2)$$