A thing to remember is that the customer is interested in the probability of correct reception after the error-correcting-code has done its magic.

The 5-dimensional Reed-Muller code of length 16 and minimum Hamming distance is capable of correcting 3 errorneous bits. The 6-dimensional Reed-Muller code of length 32 and minimum Hamming distance 16 is capable of correcting 7 errorneous bits. Let us assume a very simplistic model in which all the bits are received errorneously at the same probability $p\in(0,1)$, independently from each other. The probability of correctly decoding a received word of $R(4,1)$ is thus

$$P_4(p)=\sum_{i=0}^3{16\choose i}p^i(1-p)^{16-i},$$

and the same probability with the code $R(5,1)$ is

$$P_5(p)=\sum_{i=0}^7{32\choose i}p^i(1-p)^{32-i}.$$

Unless I made mistake, we have $P_5(p)>P_4(p)$ for most small values of $p$.
For example, $P_5(0.1)=0.9883$ and $P_4(0.1)=0.9316$. And at $p=0.01$ we have
$1-P_5(0.01)=8.5\cdot10^{-10}$ and $1-P_4(0.01)=1.7\cdot10^{-5}$.

So when transmitting an image of 1000 x 1000 pixels at $p=0.01$, we expect to receive an image free of errors, when using $R(5,1)$, but expect a few dozen garbled pixels, when using $R(4,1)$. Furthermore, to correctly receive 5 pixels worth of image data, we need to correctly receive 5 blocks of $R(5,1)$ instead of 6 blocks of $R(4,1)$. In other words, in terms of payload the fair comparison should be made between $P_4(p)^6$ and $P_5(p)^5$.

Above I assumed a decoding logic decoding up to the guaranteed error-correction probability only. One might attempt a more complicated receiver (using soft input) doing full soft decision decoding (which in this case amounts to a simple Wals-Hadamard transformation). I don't know whether that would change the verdict, though.

Moral: long codes often work better. The reason is that in a short block the number of errorneous bits has a higher (relative) variance, and thus it is easier for the number of errors to exceed the error-correcting-capability of the code.