Given first-order logic with equality and the real field $\mathbb{R} = (R, 0, 1, <, +, \cdot)$, is $\pi$ first-order definable?
By first-order definable, I mean a sentence of the form $\exists x \;\phi(x)$ such that $\pi$ is the only element in $R$ satisfying $\phi$.
No, you cannot. Although $\pi$ has "reasonably concrete" definitions in terms of $+, \times, <$ (e.g. via infinite series), none of them can be made first-order. This follows, e.g., from the fact that:
The algebraic reals form a real closed field.
The theory of real closed fields is complete, and in fact if $F_1, F_2$ are real closed fields with $F_1\subseteq F_2$, then $F_1\preccurlyeq F_2$.
$\pi$ is not algebraic.