Probably the easiest way to get an arbitrary precision Mandelbrot set using full-precision integers is to combine two integers into a rational number $p/q$. These numbers are added by finding the common denominator:
$$ \frac{m}{n} + \frac{p}{q} = \frac{mq + np}{nq} $$
and multiplied by just multiplying the respective numerators and denominators:
$$ \frac{m}{n} * \frac{p}{q} = \frac{mp}{nq} $$
Then a single point $\vec{p}$ would be represented in rational coordinates by
$$ \vec{p} = \left( \frac{n_x}{d_x} , \frac{n_y}{d_y} \right) $$
and the iteration for the point would proceed
\begin{eqnarray*}
\vec{z} &\rightarrow& \vec{z}^2 + \vec{p} \\
\left( \frac{a_x}{b_x} , \frac{a_y}{b_y} \right) &\rightarrow&
\left( \frac{a_x^2}{b_x^2} - \frac{a_y^2}{b_y^2}
, \frac{2 a_x a_y}{b_x b_y} \right) +
\left( \frac{n_x}{d_x} , \frac{n_y}{d_y} \right)\\
&=&
\left( \frac{a_x^2 b_y^2 - a_y^2 b_x^2}{b_x^2 b_y^2} +
\frac{n_x}{d_x}
, \frac{2 a_x a_y}{b_x b_y} + \frac{n_y}{d_y} \right) \\
&=& \left(
\frac{(a_x^2 b_y^2 - a_y^2 b_x^2)d_x + n_x b_x^2 b_y^2}
{b_x^2 b_y^2 d_x},
\frac{2 a_x a_y d_y + b_x b_y n_y}{b_x b_y d_y}
\right).
\end{eqnarray*}
Easy this may be, but I can't claim it'll be fast; near the boundaries of the Mandelbrot set, the numerators and denominators in the iteration will get pretty big fairly soon, even if you reduce the rationals by dividing by the greatest common denominator.