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Let $z = x + iy$, where $x, y \in \mathbb{R}$.

The given equation is:
$|z + 1| = z + 2(1 + i)$

Substitute $z = x + iy$ into the equation:
$|(x + 1) + iy| = (x + iy) + 2 + 2i$
$\sqrt{(x + 1)^2 + y^2} = (x + 2) + i(y + 2)$

Since the left-hand side is purely real, the imaginary part of the right-hand side must be zero:
$y + 2 = 0 \implies y = -2$

Equating the real parts:
$\sqrt{(x + 1)^2 + y^2} = x + 2$

Substitute $y = -2$:
$\sqrt{(x + 1)^2 + (-2)^2} = x + 2$
$\sqrt{x^2 + 2x + 5} = x + 2$

Squaring both sides (with condition $x + 2 \ge 0$):
$x^2 + 2x + 5 = (x + 2)^2$
$x^2 + 2x + 5 = x^2 + 4x + 4$
$2x + 5 = 4x + 4$
$2x = 1 \implies x = \frac{1}{2}$

Since $x = \frac{1}{2} \ge -2$, the solution is valid.

Therefore, the value of $z$ is:
$z = \frac{1}{2} - 2i$