My try:

@below: I agree, Thx

iniffur wrote: My try:

It is kinda true but you have to prove there isn't any solution from other cases From vieta $x_1+x_2+x_3=-9$ and $x_1x_2x_3=11+2^y$ So there are 2 negative root so that, there is 1 solution for $x$

We can rewrite it as $(x+3)^3 = 2^y + 38(x+1)$. So $2^{6y}\equiv 1 \pmod{19}$ and by Fermat's theorem we get \[ 2^{(6y,18)}\equiv 1\pmod{19}. \]If $y$ is not divisible by $3$, then we get a contradiction. Therefore, we must have $y\equiv 0 \pmod{3}$. Let $y=3z$ for some positive integer $z$. Then $(x+3)^3-38(x+1) = (2^z)^3=A^3$, and we get $A<x+3$, i.e. $A\le x+2$. On the other hand, clearly $2\le x$ and $x<A$, i.e. $x+1\le A$. Hence $A=x+1$ or $A=x+2$. By checking, we find $A=x+1$ has a solution $x=3$, $y=6$.

Write $(x+3)^3 - 38(x+1) = 2^y$ and consider mod $19$. Since $19$ does not divide $x+3$ (otherwise it would divide $2^y$), from the congruence $(x+3)^3 \equiv 2^y \pmod {19}$ we obtain $(x+3)^6 \equiv 2^{6y} \pmod {19}$ and hence $7^y \equiv 1 \pmod {19}$ by Fermat's Little Theorem. Since $7^1 \equiv 7$, $7^2 \equiv 11$, $7^3 \equiv 1$, it is necessary that $y$ is divisible by $3$; in particular $(x+3)^3 - 38(x+1)$ must be a perfect cube. Now since $(x+3)^3 - 38(x+1) < (x+3)^3$, we obtain $(x+3)^3 - 38(x+1) \leq (x+1)^3$ or $(x+3)^3 - 38(x+1) = (x+2)^3$. The latter is as $3x^2 - 23x - 19 = 0$, with discriminant $D = 23^2 + 4 \cdot 3 \cdot 19 = 757$, which is not a perfect square as it ends in $7$. The former requires $3x^2 - 7x \leq 6$, which is false for $x\geq 4$ as $3x^2 - 7x = x(3x-7) \geq 4 \cdot 5 > 6$. Finally, if $x=1$, the expression in the original equation equals $-12$, for $x=2$ it is $11$ and for $x=3$ it is $64 = 2^6$. (One can skip $x=2$ by noting from the beginning that $x$ must be odd.) Therefore the only solution is $(x,y) = (3,6)$.

Very Easy Assume $y >3 \implies 2^y \equiv 0 (\mod 8)$ $2^y=x^3+9x^2-11x-11 \equiv -2(x+1) (\mod 8) \implies -2(x+1) \equiv 0 (\mod 8) \implies x \equiv 3,7 (\mod 8)$ $\textbf{Case 1.}$ $x \equiv 7(\mod 8)$ $(x-7)(x^2+16x+101)=2^y -696$. We know that $v_2(696)=3$ so we get the following: $v_2( (x-7)(x^2+16x+101))<=3$ but $v_2(x-7)>=3 , v_2(x^2+16x+101)=1 \implies v_2( (x-7)(x^2+16x+101))>=4 \rightarrow \leftarrow$. Now checking for $ y \in {3,2,1}$ gives us no solution. $\textbf{Case 2.}$ $x \equiv 3(\mod 8)$ $(x-3)(x^2+12x+25)=2^y-64$ Now if $x \not\equiv 3(\mod 64)$ then $v_2(x^2+12+25)=1 \implies v_2(x-3) <6 \implies y \leq 5$ which gives us no solution. So $x \equiv 3 (\mod 64) \implies v_2((x-3)(x^2+12x+25)) \geq 7$ but we have $v_2(2^y-64) \leq 7$ Combining with $(x-3)(x^2+12x+25)=2^y-64$ we get that: $v_2(2^y-64)=7 \implies y=6 , x=3$ So $(x,y)=(3,6) \blacksquare$ Any extra cases that may have not been written are clearly trivial. Thanks for the latex ItsBesi