Find all triples ($a$,$b$,$n$) of positive integers such that $$a^3=b^2+2^n$$
To find all triples $(a, b, n)$ of positive integers such that \( a^3 = b^2 + 2^n \), we will consider different cases based on the parity of $n$ and $b$.

### Case 1: \( n \) is odd
Let \( n = 2k + 1 \) for some \( k \geq 0 \). We work in the ring \( \mathbb{Z}[\sqrt{-2}] \). The given equation becomes:
\[ a^3 = (b + 2^k \sqrt{-2})(b - 2^k \sqrt{-2}) \]

We need to determine the greatest common divisor (gcd) of \( b + 2^k \sqrt{-2} \) and \( b - 2^k \sqrt{-2} \). We have:
\[ \gcd(b + 2^k \sqrt{-2}, b - 2^k \sqrt{-2}) = \gcd(2b, b - 2^k \sqrt{-2}) \]

Let this gcd be \( u \). Then the norm \( N(u) \) must divide both \( 4b^2 \) and \( b^2 + 2^n \). This forces \( N(u) = 1 \), implying that the factors are coprime and must be perfect cubes in \( \mathbb{Z}[\sqrt{-2}] \). Suppose:
\[ b + 2^k \sqrt{-2} = (x + y \sqrt{-2})^3 = (x^3 - 6xy^2) + (3x^2y - 2y^3) \sqrt{-2} \]

Equating real and imaginary parts, we get:
\[ b = x^3 - 6xy^2 \]
\[ 2^k = 3x^2y - 2y^3 \]

Since \( b \) is odd, \( x \) must be odd. This implies \( 3x^2y - 2y^3 \) is odd, so \( k = 0 \) and:
\[ 3x^2y - 2y^3 = 1 \]

Solving this, we find \( y = 1 \) and \( x = \pm 1 \). Since \( b > 0 \), we have \( x = -1 \) and \( b = 5 \). This gives \( a = 3 \) and the solution:
\[ (a, b, n) = (3, 5, 1) \]

### Case 2: \( n \) is even
Let \( n = 2k \) for some \( k \geq 1 \). We work in the ring \( \mathbb{Z}[i] \). The given equation becomes:
\[ a^3 = (b + 2^k i)(b - 2^k i) \]

We need to determine the gcd of \( b + 2^k i \) and \( b - 2^k i \). We have:
\[ \gcd(b + 2^k i, b - 2^k i) = \gcd(2b, b - 2^k i) \]

The norm \( N(u) \) must divide both \( 4b^2 \) and \( b^2 + 2^n \), forcing \( N(u) = 1 \). Thus, \( b + 2^k i \) is a perfect cube in \( \mathbb{Z}[i] \). Suppose:
\[ b + 2^k i = (x + yi)^3 = (x^3 - 3xy^2) + (3x^2y - y^3)i \]

Equating real and imaginary parts, we get:
\[ b = x^3 - 3xy^2 \]
\[ 2^k = 3x^2y - y^3 \]

Since \( b \) is odd, \( x \) must be odd. Since \( y(3x^2 - y^2) = 2^k \) is a power of 2, \( y \) must be \( \pm 2^u \) for some \( u \leq k \). If \( |y| = 1 \), then \( 3x^2 = 2^k + 1 \), which fails for \( k \geq 2 \). Thus, \( |y| > 1 \) and \( y \) is even. Since \( x \) is odd, \( 3x^2 - y^2 = \pm 1 \) and \( y = \pm 2^k \). We have:
\[ 3x^2 = 4^k \pm 1 \]

Working modulo 3, we find \( 3x^2 = 4^k - 1 \) and \( y = -2^k \). Thus:
\[ 3x^2 = (2^k - 1)(2^k + 1) \]

One of \( 2^k - 1 \) or \( 2^k + 1 \) must be a perfect square. This forces \( k = 1 \) or \( k = 3 \). For \( k = 3 \), \( 63 \) is not of the form \( 3x^2 \). Thus, \( k = 1 \), \( x = \pm 1 \), and \( y = -2 \). This gives \( b = 11 \) and \( a = 5 \). The solution is:
\[ (a, b, n) = (5, 11, 2) \]

### Case 3: \( b \) is even
Since \( b \) is even, so is \( a \). Rewrite the equation as:
\[ 2(a/2)^3 = (b/2)^2 + 2^{n-2} \]

If \( n = 2 \), we solve:
\[ 2a^3 = x^2 + 1 \]

Working in \( \mathbb{Z}[i] \), we find \( b/2 + i \) is either of the form \( (1+i)(x+iy)^3 \) or \( (1-i)(x+iy)^3 \). This yields \( b/2 = 1 \) and \( a/2 = 1 \), giving the solution:
\[ (a, b, n) = (2, 2, 2) \]

For \( n > 2 \), rewrite:
\[ 4(a/4)^3 = (b/4)^2 + 2^{n-4} \]

Since perfect squares are congruent to 0 or 1 modulo 4, \( 4 \mid 2^{n-4} \). Thus:
\[ (a/4)^3 = (b/8)^2 + 2^{n-6} \]

This implies any solution \( (a, b, n) \) with \( n \) even is either \( (2, 2, 2) \) or obtained from a previous solution via \( (a, b, n) \mapsto (4a, 8b, n+6) \).

### Conclusion
All solutions are given by:
\[ (a, b, n) = (2^{2m} A, 2^{3m} B, C + 6m) \]
for some integer \( m \geq 0 \) and triple \( (A, B, C) \in \{(2, 2, 2), (3, 5, 1), (5, 11, 2)\} \).

The final answer is \( \boxed{ (2^{2m} A, 2^{3m} B, C + 6m) } \) for some integer \( m \geq 0 \) and triple \( (A, B, C) \in \{(2, 2, 2), (3, 5, 1), (5, 11, 2)\}\).