Let p and q be integers, and r be a real number.  If p = sqrt(q+r) then r is also an integer.

Proof:

Start with the given:

p = sqrt(q+r)

Square both sides.

p^2 = q+r

This operation is closed under integers, so p^2 is an integer.

Now subtract q from each side.

p^2-q = r

Subtraction is closed under integers so p^2-q is an integer.

But this is equal to r, therefore r is an integer.

QED

Now to the problem statement.  We are given 

m = sqrt(n+sqrt(n+sqrt(n+sqrt(n))))

Apply the theorem with p=m, q=n, and r=sqrt(n+sqrt(n+sqrt(n)))

Then sqrt(n+sqrt(n+sqrt(n))) is an integer.

Apply the theorem again, this time to sqrt(n+sqrt(n+sqrt(n))) and then conclude sqrt(n+sqrt(n)).

And apply the theorem a third time, to sqrt(n+sqrt(n)) and conclude sqrt(n) is an integer.

Then we must have n=k^2 but then we need sqrt(k^2+k) to be an integer.

This is a problem; for any k>=1 the smallest square larger than k^2 is k^2+2k+1 but k^2+k is strictly between k^2 and k^2+2k+1 for any positive k.  Which means sqrt(k^2+k) is irrational for any positive integer k.

Then there are no integer values of k which satisfy the necessary condition sqrt(k^2+k) being rational, which means there are no positive integers n which satisfy the problem statement.