2. Solve the equation in natural numbers: $1+x+x^{2}+x^{3}=2^{y}$.
2. Transform the equation to the form: $\left(1+x^{2}\right)(1+x)=2^{y}$, from which $1+x^{2}=2^{m}$, where $m$ is a non-negative integer, and $1+x=2^{n}$, where $n$ is a non-negative integer. Since $x=2^{n}-1$, then $x^{2}=\left(2^{n}-1\right)^{2}=2^{2 n}-2 \cdot 2^{n}+1$. Considering that $1+x^{2}=2^{m}$, we get $2^{2 n}-2 \cdot 2^{n}+2=2^{m}$. Dividing both sides of the equation by 2, we obtain: $2^{2 n-1}$ $-2^{n}+1=2^{m-1}$, which is equivalent to the equation $2^{n}\left(2^{n-1}-1\right)+1=2^{m-1}$. Since $m$ is a non-negative integer, for $m>1$, $2^{m-1}$ is an even number, so $2^{n}\left(2^{n-1}-1\right)$ must be an odd number. This is impossible. Therefore, we need to consider $m=0$ and $m=1$. For $m=0$, we have $1+x^{2}=2^{0}$, from which $x=0, y=0$. And for $m=1$, we get $x^{2}=1$ and, considering that $x$ is a natural number, we obtain $x=1, y=2$.