We can't use Rouche's theorem here directly, so we have to apply the argument principle. If $f(z) = \tan(\pi z) - z$ , then $f'(z) = \pi \sec^2(\pi z) - 1$. Choose the rectangle $\Gamma$ with endpoints $(1- \varepsilon) \pm Ri$ and $(-1 + \varepsilon) \pm Ri$, and evaluate
$$\frac{1}{2 \pi i}\oint_\Gamma \frac{f'(z)}{f(z)} dz = \frac{1}{2 \pi i}\oint_\Gamma \frac{\pi \sec^2(\pi z) - 1}{\tan(\pi z) - z} dz,$$
and let $R \to \infty$.
Notice that the integrand is odd, so the integral evaluates to zero (right?). Therefore, by the argument principle $f$ has as many poles as it does zeroes. But $f(z)$ has no poles in $|\Re (z)| < 1$.
What am I missing here?
Thanks to Daniel for pointing out my obvious mistake.
Pick the branch of the logarithm with argument in $[-\pi,\pi]$. Integrate on the rectangle from the point $-1 + \varepsilon_1 - i\varepsilon_2$ to $-1 + \varepsilon_1 + i\varepsilon_2$, following $\gamma$ counterclockwise. Since $\log(f) = f'/f$ on this section of the curve, letting $\varepsilon_1 \to 0$ and setting $\varepsilon = \varepsilon_2$, the integral comes out to
$$\begin{align} &\log[\tan(-\pi + \pi i\varepsilon) - (-1 + i\varepsilon)] - \log[\tan(-\pi - \pi i\varepsilon) - (-1 - i\varepsilon)]\\ \rightarrow & (\log|-1| + i\pi) - (\log |-1| - i \pi) \\ = &2\pi i, \end{align}$$
which is unchanged as $R \to \infty$. By the argument principle, $f(z)$ has one zero more than it has poles. Since $\tan(\pi z) - z$ has poles at $z = \pm \frac{1}{2}$, $f(z)$ has three zeroes in the strip, and $\tan(\pi z) = z$ has exactly three solutions in the strip $|\Re(z)| < 1$.