A fast (and correct) way of finding a common divisor for $1805$, $1991$, $2146$ so that they have the same remainder

If the remainders from $1991$ and $1805$ are the same, then these numbers are congruent mod $n$, so $n$ divides $1991-1805=186$. Similarly, $n$ must divide $2146-1991=155$. So you need a number that divides both $186$ and $155$, so it will divide their difference: $186-155=31$. Since $31$ is prime its only divisors are $31$ and $1$, and technically the answer is both of these.

So your method is valid. And very fast, so I wouldn't expect anything faster.

Given three numbers $a,b,c$ your are looking for $\gcd (b-a,c-b,c-a)$. Your approach is a good step and the answer is guaranteed to divide $b-a$ and $c-b$. You have done one step in Euler's method, but the answer could be smaller. For example, let the numbers be $1,19,33$. You would do $19-1=18, 33-19=14, 18-14=4$ but are not done as the $\gcd$ is $2$. You can now keep going with $14-3 \cdot 4=2$ and you are done because $2$ divides both $18$ and $14$.