Does Reed-Muller codes exist when $r = 0$ and $m = 0$?
The codewords in a Reed-Muller code of degree $d$ and length $2^n$ are binary vectors of the form
$$\big(f(0,0,\ldots, 0, 0), f(0,0, \ldots, 0, 1),
f(0,0, \ldots, 1, 0), f(0,0, \ldots, 1, 1), \cdots, f(1,1,\ldots, 1,1)
\big)$$
where $f$ is the corresponding binary polynomial of degree at most
$d$ in $n$ variables. 

Suppose first that $n &gt; 0$. The dimension of the code is
$$k = \sum_{i=0}^d \binom{n}{i}\tag{1}$$
If $d=0$, then $k = \binom{n}{0}=1.$ There are only two "polynomials"
of degree $0$ in $n$ variables, namely the constants $0$ and $1$, and consequently
the codewords of length $2^n$ are $000\cdots 0$ and $111\cdots 1$.

If $n=0$ also, the codewords (if any) are of length $2^0 = 1$.
Now, you might want to claim from $(1)$ that $k = \binom{0}{0} = \frac{0!}{0!(0-0)!} = 1$ and so there are $2^k = 2$ codewords 
(which, of course, are $0$ and $1$).
Alternatively, the $k$ in Equation $(1)$ is the total number of
subsets of cardinality $d$ or less of a set of $n$ elements.
There is exactly one subset (of cardinality $0$) 
of the set of $0$ elements (namely,the empty set $\emptyset$).
and that subset is $\emptyset$ itself. Thus, the Reed-Muller code
of order $0$ and length $2^0$ is the $(1,1)$ code consisting
of two codewords $0$ and $1$. As a linear code, this is the
identity map.