1. Base Case Verification:
We start by verifying the base case for $j = 0$.
$$L_{2^0} = L_1 = 1$$
$$1 + L_{2^0} = 1 + 1 = 2 \equiv 0 \pmod{2^{0+1}} = 2$$
Thus, the base case holds.

2. Inductive Hypothesis:
Assume that the statement is true for some $j \geq 0$, i.e.,
$$1 + L_{2^j} \equiv 0 \pmod{2^{j+1}}$$
This means there exists an integer $m$ such that:
$$L_{2^j} = 2^{j+1}m - 1$$

3. Inductive Step:
We need to show that the statement holds for $j+1$, i.e.,
$$1 + L_{2^{j+1}} \equiv 0 \pmod{2^{j+2}}$$
Using the given identity:
$$L_{2^{k+1}} + 1 = L_{2^k}^2 - 1$$
Set $k = j$:
$$L_{2^{j+1}} + 1 = L_{2^j}^2 - 1$$
Substitute the inductive hypothesis $L_{2^j} = 2^{j+1}m - 1$:
$$L_{2^{j+1}} + 1 = (2^{j+1}m - 1)^2 - 1$$
Expand the square:
$$(2^{j+1}m - 1)^2 = 2^{2j+2}m^2 - 2 \cdot 2^{j+1}m + 1 = 2^{2j+2}m^2 - 2^{j+2}m + 1$$
Therefore:
$$L_{2^{j+1}} + 1 = 2^{2j+2}m^2 - 2^{j+2}m + 1 - 1 = 2^{2j+2}m^2 - 2^{j+2}m$$
Factor out $2^{j+2}$:
$$L_{2^{j+1}} + 1 = 2^{j+2}(2^{j}m^2 - m)$$
Since $2^{j+2}$ is a factor, we have:
$$L_{2^{j+1}} + 1 \equiv 0 \pmod{2^{j+2}}$$
This completes the inductive step.

By mathematical induction, we have shown that $1 + L_{2^j} \equiv 0 \pmod{2^{j+1}}$ for all $j \geq 0$.

$\blacksquare$