How to prove order of growth in Big Theta

This is an algorithm of BruteForceStringMatch found from the book "The design and analysis of algorithm" by anany Levitin at pg105.
The description of the algorithm:
Given a string of characters called the text and a string of m characters(m<=n) called the pattern, find the starting index of substring of the text that

matches the pattern.

Followed is the pseudo code:
for i<-0 to n-m do
  j <- 0
  while j<m and p[j]=T[i+j]
    j<-j+1
  if j=m return i
return -1

The book says that for this algorithm, the order of growth in Big Theta is nm, but it doesn't give the steps of proving it.

Hence I tried to write the prove on my own, this is what I try:
Since the worst case is to make all m comparisons before shifting the pattern and this can be happen for each of the n-m+1 tries, hence is has a total of m

(n-m+1) tries
To prove m(n-m+1) belong to the order of growth in Big Theta is nm, we need to show for all n>=n0, m>=m0 that
a(mn)<=m(n-m+1)<=b(mn) where a and b are positive integers and n0, m0 are nonnegative integers.
To prove the right inequality(upper bound)
m(n-m+1)
<=m(n-m+1+(m-1)) when m>=1,n>=0
<=mn
This proved the right inequality.(with b=1, m>=1 ,n>=0)

Is the assumption and procedure for proving right inequality correct? I am not sure of it because the book only mentioned for function of one input

parameter, says t(n), to show that the order of growth in Big Theta is g(n) we need to prove a(g(n))<=t(n)<=b(g(n)) for all n>=n0.
But this question seems like involved two input parameters. Is it correct that for function of two input parameters, says t(n,m), to show that the order of

growth in Big Theta is g(n,m) we need to prove a(g(n,m))<=t(n,m)<=b(g(n,m)) for all n>=n0, m>=m0?

If it is correct, may I know how to prove the left inequality, ie. a(mn)<=m(n-m+1)?