Such a painful case bash (for me atleast). $WLOG x\geq y\geq z\geq t$ so $xyz+yzt+xzt+xty\leq 4xyz$ so $4xyz\geq xyzt +3$ so $t<4$. 1.case $t=1$ so $xy+yz+xz=3$ so $x=y=z=t=1$. 2.case $t=2$ so $xyz +3=2\cdot (xy+xz+yz)$, again $2\cdot (xy+xz+xy)\leq 6xy$, so $xyz+2\leq 6xy$ so $z<6$ 1.subcase $z=5$ so $3xy=10x+10y-3$ $10(x+y)\leq 20x$ so $xy\leq \frac{20}{3}\cdot x-1$, because $y\geq z$ so try $y=5$ which don't have solutions. 2.subcase $z=4$, obviously $LHS$ is odd and $RHS$ is even. 3.subcase $z=3$ we get $xy=6x+6y-3$ as in previous cases $y\leq 9$, so aply $y\in {7,8,9}$,$x=39,y=7,z=3,t=2$,$x=17,y=9,z=3,t=2$ 4.subcase $z=2$ we get obviously there is no solutions. 3.case $t=3$ we get $3(xy+yz+xz)=2xyz+3$ so we easly get $LHS\leq 9xy$ so $2xyz+3\leq 9xy$ so $z\leq 4$ 1.subcase $z=4$ we have $12(x+y)=5xy+3$ so we get $y\leq 4$ so we yust need to check $y=4$ which isn't solution. 2.subcase $z=3$ we have $xy+1=3x+3y$ so check $y\in{3,4,5}$ we get $t=3,z=3,y=4,x=11$, $t=2,z=3,y=3,x=7$. So finaly solutions: ${x=y=z=t=1}$ ,${x=39,y=7,z=3,t=2}$ ,${x=17,y=9,z=3,t=2}$ ,${x=11,y=4,z=3,t=3}$ ,${x=7,y=3,z=3,t=2}$.

Garfield wrote: Such a painful case bash (for me atleast). $WLOG x\geq y\geq z\geq t$ so $xyz+yzt+xzt+xty\leq 4xyz$ so $4xyz\geq xyzt +3$ so $t<4$. 1.case $t=1$ so $xy+yz+xz=3$ so $x=y=z=t=1$. 2.case $t=2$ so $xyz +3=2\cdot (xy+xz+yz)$, again $2\cdot (xy+xz+xy)\leq 6xy$, so $xyz+2\leq 6xy$ so $z<6$ 1.subcase $z=5$ so $3xy=10x+10y-3$ $10(x+y)\leq 20x$ so $xy\leq \frac{20}{3}\cdot x-1$, because $y\geq z$ so try $y=5$ which don't have solutions. 2.subcase $z=4$, obviously $LHS$ is odd and $RHS$ is even. 3.subcase $z=3$ we get $xy=6x+6y-3$ as in previous cases $y\leq 9$, so aply $y\in {7,8,9}$,$x=39,y=7,z=3,t=2$,$x=17,y=9,z=3,t=2$ 4.subcase $z=2$ we get obviously there is no solutions. 3.case $t=3$ we get $3(xy+yz+xz)=2xyz+3$ so we easly get $LHS\leq 9xy$ so $2xyz+3\leq 9xy$ so $z\leq 4$ 1.subcase $z=4$ we have $12(x+y)=5xy+3$ so we get $y\leq 4$ so we yust need to check $y=4$ which isn't solution. 2.subcase $z=3$ we have $xy+1=3x+3y$ so check $y\in{3,4,5}$ we get $t=3,z=3,y=4,x=11$, $t=2,z=3,y=3,x=7$. So finaly solutions: ${x=y=z=t=1}$ ,${x=39,y=7,z=3,t=2}$ ,${x=17,y=9,z=3,t=2}$ ,${x=11,y=4,z=3,t=3}$ ,${x=7,y=3,z=3,t=2}$. Unfortunately I doubt there's a better way, as the official solution is more or less as yours, bounding the variables and checking all the cases and subcases. To be fair I don't like this year's problem, they are all brute-force

Also $(x,y,z,t) = (3,1,1,0)$ if your definition of $\mathbb{N}$ includes $0$.