(2^(N+1))-1
Suppose level is 2 then total number of nodes will be
1 root
2 left of root and right of root
2 left and right of left of root
2 left and right of right of root
so total nodes are 1+2+2+2=7

to be more generic this kind of problem is best solved
recursivly.

To point out that 2 ^ N AND 3 ^ N are both wrong,
here's a few examples: (the exponet is the amount of levels)
2^0 = 1, correct
2^1 = 2, incorrect, should be 3
2^2 = 4, incorrect, should be 7

And a tree with three children
3^0 = 1, correct
3^1 = 3, incorrect, should be 4
3^2 = 9, incorrect, should be 13

Looking at that I'm sure you can see the pattern.
Let
C = "Number of Possible Children"
N = Levels

N
Σ C^N
j=0

or in C++ code
int NodeCount(int C, int N)
{
if (N < 0) return 0
return NodeCount(C, N-1) + C^N
}

If root is at level 0 then :
Case 0:
When level is 1 max nodes is 1
Case 1:
When level is 1 then max would be 3.
Case 2:
When level is 2 then max nodes would be 7

because root is at level 0. and there are n levels only.
so last level i.e level of leaves should be n-1. for maximum
we will consider complete binary tree which is full at level
n-1.

for 0 level -> 2^0 i.e 1 element
for 1 level -> 2^1
...
so on

for n-1 levle ->2^(n-1) nodes
---------------------------
sum = 2^n-1