What might I use to show that an entire function with positive real parts is constant?

The other three answers are overkill to me.. Simply consider $e^{-f}$ if $f$ is your function. Is it bounded?

It isn't a nonconstant polynomial, by the fundamental theorem of algebra. It doesn't have an essential singularity at infinity, by the Casorati-Weierstrass theorem. What other possibilities are there?

Alternatively, if you add $1$, you get a function satisfying $|g(z)|\geq 1$ for all $z$. What can you say about the reciprocal of $g$?

Well, can't we just say that, since $-f(z)$ is entire, $e^{-f(z)}$ is also entire, and if we write

$f(z) = u(z) + iv(z), \tag{1}$

where $u(z)$, $v(z)$ are the (harmonic) real and imaginary parts of $f(z)$ (so that $u(z) = Re \; f(z)$), then

$\vert e^{-f(z)} \vert = \vert e^{-u(z) - iv(z)} \vert = \vert e^{-u(z)} \vert \vert e^{-iv(z)} \vert = e^{-u(z)}, \tag{2}$

since

$e^{-u(z)} > 0 \tag{3}$

and

$\vert e^{-iv(z)} \vert = \vert \cos (-v(z)) + i\sin (-v(z)) \vert = 1; \tag{4}$

but $-u(z) < 0$ by hypothesis; thus $e^{-u(z)} < 1$, whence $e^{-f(z)}$ is a bounded entire function, hence constant; hence $f(z)$ must itself be a constant. QED.

Can't we just say that? I think we can!