Since $3^{x} +x^{4} =y!+2019 >2019$, we have $x\geq 6$. Assume that $y\geq 6$, then $x^{4} \equiv 3\bmod 9$, impossible. So we must have $y\leq 5$. Assume that $x\geq 7$, then $3^{7} +7^{4} \leq y!+2019\leq 2019+120$, impossible. So we must have $x=6$, and so $y=3$. Hence the answer is $( x,y) =( 6,3)$.

minecraftfaq wrote: Find all positive intengers $x,y$, satisfying: $$3^x+x^4=y!+2019.$$

Suppose that $y \geq 4$, then clearly $x$ is even. Checking $(mod.8)$ we have that $LHS=1(mod.8), RHS=3(mod.8)$, contradiction. Therefore $1 \leq y \leq 3$, and computation gives $(x,y)=(6,3)$

We can see that if $y=1,2 $ there are no solutions . So $ y \geq 3 $ and this implies that $ 6 \mid x $ Assume that $ y \geq 7 $ take mod $7$ amd we have contradiction So $ 6 \geq y $ and we can check that only solution is $(x,y)=(6,3)$