Hint $ $ Uniqueness of radix rep can be deduced intuitively from the simple fact that an integer root of an integer coef polynomial divides the least degree coef (i.e. Rational Root Test). For example
$\qquad11001_2 = g(2),\,\ \ g(x) = x^4+x^3+1$
$\qquad 10011_2 = h(2),\ \ h(x) = x^4+x+1$
If they're equal $\, 0 = g(2)-h(2) =: f(2)\,$ for $\,f = g-h = x(x^2-1)\,$ so $\,2\,$ is a root of $\,x^2-1\,$ so $\, 2^2 = 1\,\Rightarrow\, 2\mid 1,\,$ contradiction. This idea works generally - the nonzero coef's of $g-h$ are $\pm1$ contra the root $2$ must divide the least degree such coef. Below is the proof for general radix.
If $\,g(x) = \sum g_i x^i$ is a polynomial with integer coefficients $\,g_i\,$ such that $\,0\le g_i < b\,$ and $\,g(b) = n\,$ then we call $\,(g,b)\,$ a radix $\,b\,$ representation of $\,n.\,$ It is unique: $ $ if $\,n\,$ has another rep $\,(h,b),\,$ with $\,g(x) \ne h(x),\,$ then $\,f(x)= g(x)-h(x)\ne 0\,$ has root $\,b\,$ but all coefficients $\,\color{#c00}{|f_i| < b},\,$ contra the below slight generalization of: $ $ integer roots of integer polynomials divide their constant term.
Theorem $\ $ If $\,f(x) = x^k(\color{#0a0}{f_0}\!+f_1 x +\cdots + f_n x^n)=x^k\bar f(x)\,$ is a polynomial with integer coefficients $\,f_i\,$ and with $\,\color{#0a0}{f_0\ne 0}\,$ then an integer root $\,b\ne 0\,$ satisfies $\,b\mid f_0,\,$ so $\,\color{#c00}{|b| \le |f_0|}$
Proof $\ \ 0 = f(b) = b^k \bar f(b)\,\overset{\large b\,\ne\, 0}\Rightarrow\, 0 = \bar f(b),\,$ so, subtracting $\,f_0$ from both sides yields $$-f_0 =\, b\,(f_1\!+f_2 b+\,\cdots+f_n b^{n-1})\, \Rightarrow\,b\mid f_0\, \overset{\large \color{#0a0}{f_0\,\ne\, 0}}\Rightarrow\, |b| \le |f_0|\qquad {\bf QED}\qquad\quad$$
Remark $\ $ Thus uniqueness of radix rep is essentially a special case of the Rational Root Test,