(1) Proof: Let's assume $x_1 > x_2$ (where $x_1, x_2 \in \mathbb{R}$), then $x_1 - x_2 > 0$. Since for $x > 0$, we have $f(x) > 1$,
it follows that $f(x_1) - f(x_2) = f((x_1 - x_2) + x_2) - f(x_2) = f(x_1 - x_2) + f(x_2) - 1 - f(x_2) = f(x_1 - x_2) - 1 > 1 - 1 = 0,$
hence $f(x_1) > f(x_2)$,
which confirms that $f(x)$ is an increasing function on $\mathbb{R}$.

(2) Since $f(x)$ is an increasing function on $\mathbb{R}$, from the inequality $f( \sqrt {m}) + f( \sqrt {m} \cdot x) > f(x^{2} - 1) + 1$, we can infer that
$f((1 + x)\sqrt{m}) > f(x^2 - 1)$.
This implies that $(1 + x)\sqrt{m} > x^2 - 1$, and this must hold true for all $m$ such that $\frac {1}{16} \leq m \leq \frac {1}{4}$.
Let $t = \sqrt{m}$, then $t$ falls within the interval $\left[ \frac {1}{4}, \frac {1}{2} \right]$.
The original inequality is equivalent to $(1 + x)t > x^2 - 1$ for all $t \in \left[\frac {1}{4}, \frac {1}{2}\right]$.
Let $g(t) = (1 + x)t - x^2 + 1$. To ensure that $g(t) > 0$ for $t \in \left[\frac {1}{4}, \frac {1}{2}\right]$ at all times,
we need both $g\left( \frac {1}{4} \right) > 0$ and $g\left( \frac {1}{2} \right) > 0$.
Solving these inequalities yields $-1 < x < \frac {5}{4}$.
Therefore, the range for the real number $x$ is $\boxed{-1 < x < \frac {5}{4}}$.