The question is to find all entire functions that satisfy the following condition: $|f(z)| \geq |z| + 1$
My attempt:
$|f(z)| \geq |z| + 1 \geq |z + 1|$
Letting $h(z) = \frac{z+1}{ f(z)}$, $h(z)$ is bounded and entire and therefore constant by Liouville's theorem.
Then $h(z) = \frac{z+1}{ f(z)} = C$, so $f(z) = (1/C)(z+1)$, where $C$ is a constant with $|C| >1$
is this work okay?
You have correctly found that if $f$ is an entire function satisfying $\lvert f(z)\rvert \geqslant \lvert z\rvert +1$ for all $z\in \mathbb{C}$, then $f(z) = K\cdot (z+1)$ for some $K\in \mathbb{C}$. But then we have
$$\lvert f(-1)\rvert = 0 < \lvert -1\rvert + 1,$$
so such a function does not exist.
Another argument is to note that $\lvert f(z)\rvert \geqslant \lvert z\rvert + 1$ implies that $f$ has no zeros, hence $1/f$ is also an entire function, and that satisfies
$$\frac{1}{\lvert f(z)\rvert} \leqslant \frac{1}{\lvert z\rvert + 1}.$$
By Liouville's theorem, $1/f$ is constant, and since
$$\lim_{z\to \infty} \frac{1}{\lvert z\rvert + 1} = 0$$
we must have $1/f \equiv 0$. But then $f \equiv \infty$ is not an entire function.