Note that $2x+1+1 = 2(x+1)$ and $x^3 + 1 = (x+1)(x^2 - x + 1)$ with $x^2 - x + 1$ odd always. So the $\nu_2(n+1)$ goes up by $1$ when $2x+1$ is done and stays the same when you take the cube root. So the only way the numbers Alice writes are bounded is if $\nu_2(n+1)$ remains constant eventually, but then only cube roots will be taken, which cannot go on forever since $1$ will never be written on the board. $\blacksquare$

Induction for $N$ and some number theory skills.

If $Y^3$ is the first cube written written after $X^3>1$, then $2^nX+2^n-1=Y^3 \implies 2^n(X+1)=Y^3+1 =(Y+1)(Y^2-Y+1)$. Because $Y, Y^2-Y+1$ are odd, $2^n|Y+1 \implies Y^2-Y+1 |X+1 \implies X^3 >Y^3$. If $1$ is not written at any point then second move is made finite times, we are done. If $1$ is written at some point then we cannot make the second move at any time because $2^n-1 \neq x^3$ (with the same reasoning above). So we will have what we want, a large number.

The sequence goes like this $a \mapsto 2a + 1 \mapsto 4a + 3 \mapsto \cdots \mapsto 2^na + 2^n - 1$, now if \[2^na + 2^n - 1 = X^3 \iff 2^n(a+1) = (X+1)(X^2 - X + 1)\]since $X$ is odd, $\nu_2(X+1) \ge n + \nu_2(a+1)$. So for any number $t$ in the sequence, $\nu_2(t+1)$ is non decreasing (and it isn't constant) hence the sequence grows unbounded.

Thanks to $\textbf{sk2005}$ for correcting this solution. We start with $x$ and perform the operation $2x+1$ until we hit a cube. The nth term of the sequence is $2^n (x+1)-1$, say for some $n_1$th term we get a perfect cube $k_1^{3}$. Hence, $$2^n (x+1)-1 = k_1^{3} \implies 2^n (x+1)= (k_1 +1)(k_1^2 -k_1+1)$$Now $k_1$ is odd, so $k_1+1$ is even and $(k_1^2 -k_1+1)$ is odd. Hence, $$v_2 (k_1 +1)=n_1 + v_2(x+1)$$Now, as we have hit $k_1 ^3$, our next term would be $k_1$ and the sequence would continue as $$k_1^3 \rightarrow k_{1} \rightarrow 2 k_1+1 \rightarrow \ldots \rightarrow 2^{n_2}\left(k_1+1\right)-1=k_2^3$$where for some $n_2$th term, we get a cube $k_2^3$. Similarly to the previous case we will get $$\begin{aligned} v_2\left(k_2+1\right) & =n_2+v_2\left(k_1+1\right) \\ & =n_2+n_1+v_2(x+1) \end{aligned}$$So inductively, for some $t \in \mathbb{N}$ $$v_2\left(k_t+1\right)=\sum_{i=1}^t n_i+v_2(x+1)$$Hence $v_2(k_t+1)$ grows unbounded. Hence $k_t$ grows unbounded too. Note that if $k_1=j_1 ^3$ for some $j_1$, then $v_2(k_1+1)= v_2(j_1+1)$. Hence, for some $m,j$ if $2^m(j_1+1)-1=j_2^3$ then $$v_2(j_2+1)=m+v_2(j_1+1)=m+v_2(k_1+1)$$So, $v_2(j_i+1)$ grows unbounded.

The sequence goes as follows: \[N\longrightarrow2N+1\longrightarrow4N+3\longrightarrow8N+7\longrightarrow\cdots,\]unless we hit a cube. A simple induction gives that a general term is $2^KN+2^K-1$. We will prove that there is no upper limit of the sequence. Suppose there is, say $10^{100}$. Then, at each point when we hit a cube, that cube is less than $10^{100}$ and is cube-rooted. Hence, the given problem is equivalent to proving that there do not exist $2n$ positive integers $x_1,x_2,\ldots, x_n,\ k_1, k_2,\ldots, k_n$ such that \[2^{k_i}(x_i+1)=x_{i+1}^3+1\]for all $1\leq i\leq n$, where $x_{n+1}=x_1$. Suppose there do exist such $2n$ positive integers. Note that if $2^k\mid x^3+1\Longrightarrow 2^k\mid(x+1)(x^2-x+1)$. Now, $x^2-x+1$ is odd, so $2^k\mid x+1$. We have that $2^{k_{i-1}}\mid x_i+1$, and $2^{k_i}(x_i+1)\mid x_{i+1}^3+1\Longrightarrow 2^{k_{i-1}+k_i}\mid x_{i+1}+1$. A simple induction gives $2^{k_1+k_2+\cdots+k_n}\mid x_i+1$ Applying this method $t$ times gives $2^{t(k_1+k_2+\cdots+k_n)}\mid x_i+1$, which contradicts the fact that $v_2(x_i+1)$ is finite. We have obtained the required contradiction. $\blacksquare$

Neat problem. The key idea is to notice that $\nu_2(x+1)$ is always increasing (not necessarily strict) where $x$ is on the board, because, the first move strictly increases the quantity, and the second move keeps the quantity constant, due to LTE. Now, if this sequence eventually terminates, it means that $\nu_2(x+1)$ is eventually constant, which means we are continuously taking cube roots, meaning that $1$ is on the board, a contradiction. Therefore, the sequence is unbounded.

Observe that until you hit a cube, the terms you get will be of the form $2^n \cdot x+2^n-1$ therefore looking that the written number +1 in $v_2$ is worth, indeed if you ever go $x^3+1 \to x+1$ note that $v_2(x^2-x+1)=0$ therefore $v_2$ of written number +1 is constant in such case. Since taking the square root is a finite process (cuz you either hit 1 or a non-cube) eventually $v_2(N+1)$ has to increase due to $x \to 2x+1$ which adds the $v_2$ by 1. Therefore this process eventually hits a number greater or equal to $2^{1434!}>10^{100}$ thus we are done .

This was one of my favorite problems of all time Main idea is that $\nu_2(x+1)$ increases every turn where $x$ is the number on the board.