883. Solve the equation in natural numbers

$$
1+x+x^{2}+x^{3}=2^{y}
$$
$\triangle$ Obviously, $x$ is odd. Factorize the left side of the equation

$$
(1+x)\left(1+x^{2}\right)=2^{y}
$$

Then each of the numbers $1+x$ and $1+x^{2}$ must be a power of two. But for odd $x$

$$
\text { GCD }\left(1+x, 1+x^{2}\right)=2
$$

(think about why; see § 11, point 11.1.1). Since $1+x^{2} \geqslant 1+x$, the only possible case is:

$$
\left\{\begin{array}{l}
1+x=2 \\
1+x^{2}=2^{k}
\end{array}\right.
$$

From this, $x=1, k=1$, and then $y=k+1=2$.

Answer: $(1 ; 2)$.