Simple way to prove $\left \{ 0^{n}1^{m} \mid (n-m) \bmod 5=0 \right \}$ is regular?
$(n-m)\bmod 5$ can only be $0,1,2,3,4$. So intuitively, the language in the question is a regular language. We can show that conclusion rigorously in the following two ways.
Regular expression
The simplest way is to verify that the language is described by the following regular expression.
$$(00000)^*(\epsilon+01+0011+000111+00001111)(11111)^*.$$
Intuitively, the language consists of words each of which is zero or more five 0s, followed by some number of 0s, followed by the same number of 1s, followed by zero of more five 1s.

Minimal DFA
Here are the states of the minimal DFA for this language.

$Z_{0}$, which represents words each of which is zero or more five 0s.
$Z_{1}$, which represents words each of which is zero or more five 0s, followed by 0.
$Z_{2}$, which represents words each of which is zero or more five 0s, followed by 00.
$Z_{3}$, which represents words each of which is zero or more five 0s, followed by 000.
$Z_{4}$, which represents words each of which is zero or more five 0s, followed by 0000.
$S_{0}$, which represents words each of which is zero or more five 0s, followed by nothing or 01 or 0011 or 000111 or 00001111, followed by zero or more five 1s, ending with 1.
$S_{1}$, which represents words each of which is zero or more five 0s, followed by 0 or 001, or 00011 or 0000111 or 1111, followed by zero or more five 1s, ending with 1.
$S_{2}$, which represents words each of which is zero or more five 0s, followed by 00 or 0001 or 000011 or 111 or 01111, followed by zero or more five 1s, ending with 1.
$S_{3}$, which represents words each of which is zero or more five 0s, followed by 000 or 00001 or 11 or 0111 or 0001111, followed by zero or more five 1s, ending with 1.
$S_{4}$, which represents words each of which is zero or more five 0s, followed by 0000 or 1 or 011 or 00111 or 0001111, followed by zero or more five 1s, ending with 1.
$R$, which represents all other words, which are words that are not of the form $0^*1^*$.

In other words, besides the &quot;rejecting&quot; state $R$ that behaves as a black hole, we have, for $i=0,1,2,3,4$,

a state $Z_i$ which represents words of the form $0^*$ whose number of 0s is the same as $i$ modulo 5.
a state $S_i$ which represents words of the form $0^*1^*1$ whose number of 0s is the same as $i$ plus its number of 1s modulo 5.

It should be not be difficult for you to figure out the initial state, the accepting states and the transitions between the states.


Is it reasonable to get a DFA with at least 30 states for this language?

It is totally fine if you can construct a DFA with 30 states or 2019 states for this language. That will show the language is regular.
On the other hand, it is sometimes interesting or helpful or challenging to find the DFA with the minimal number of states. For this language, 11 states is the minimum.

Here are two related exercises.
Exercise 1. Show $\left \{ 0^m1^n \mid (n-m) \bmod 5=2 \right \}$ is a regular language. How many states are there in the minimal DFA for it?

Exercise 2. Show $\left \{ 0^k1^m2^n \mid (k+m-n) \bmod 2=0 \right \}$ is a regular language. How many states are there in the minimal DFA for it?