Note we need to look at cases where $v=2^s$ and $v=-2^s$. Modulo $3$ and $8$ solve them.

Again, we need to look at both cases $v=\pm 2^s$.

Was going to eat so that's why I posted an outline only

First, we solve for odd $b$ and deal with the annoying case of $b$ even later. We do casework on the parity of $n$: Case 1: $n$ is odd, say $n = 2k+1$ for some $k \geq 0$. We work in $\mathbb{Z}[\sqrt{-2}]$. We have $a^3 = (b+2^{k}\sqrt{-2})(b-2^{k}\sqrt{-2})$. Now, $\gcd(b+2^{k}\sqrt{-2}, b-2^{k}\sqrt{-2}) = \gcd(2b, b-2^k\sqrt{-2})$. Suppose this quantity is $u$. Then $N(u) \mid 4b^2$ and $N(u) \mid b^2+2^n$ which forces $N(u) = 1$. In particular, the factors of the equality are coprime and hence must be perfect cubes in $\mathbb{Z}[\sqrt{-2}]$ themselves. Suppose we have \[b+2^k \sqrt{-2} = (x + y\sqrt{-2})^3 = (x^3-6xy^2)+(3x^2y-2y^3)\sqrt{-2}\]for some $x, y \in \mathbb{Z}$. We obtain that $b = x^3-6xy^2$ and $2^k = 3x^2y-2y^3$. Observe that because $b$ is odd, $x$ is odd. It follows that $3x^2-2y^3$ is odd as well, in particular $k = 0$ and $3x^2y-2y^3 = 1$ which forces $y = 1, x = \pm 1$ after simple calculations. In particular, because $b > 0$ we obtain $x = -1$ and $b = 5$. This gives $a = 3$ and we retrieve the solution $(a, b, n) \equiv (3, 5, 1)$. Case 2: $n$ is even, say $n = 2k$ for some $k \geq 1$. This time, work in $\mathbb{Z}[i]$! We have $a^3 = (b+2^k i)(b-2^k i)$. Again, $\gcd(b+2^k i, b-2^k i) = \gcd(2b, b-2^k i)$. The norm divides both $4b^2$ and $b^2+2^n$ following which we obtain that it is $1$ and these two are coprime. In particular, $b+2^k i$ is a perfect cube in $\mathbb{Z}[i]$. Hence, we have \[b+2^k i = (x+iy)^3 = (x^3-3xy^2)+(3x^2y-y^3)i\]for some $x, y \in \mathbb{Z}$. Because $b$ is odd, we obtain that $x$ is odd. Because $y(3x^2-y^2) = 2^k$ is a perfect power of $2$, it follows that $y = \pm 2^u$ for some $u \leq k$. If $|y| = 1$, then $3x^2 = 2^k+1$ which fails for $k \geq 2$ by working mod $4$. So if $|y| = 1$ we have $x = \pm 1$ and $k = 1$. But now $b$ is even, impossible. Thus $|y| > 1$. In particular, $y$ is even. Because $x$ is odd, it follows that $3x^2-y^2 = \pm 1$ and $y = \pm 2^k$. We have $3x^2 = 4^k \pm 1$. Working mod $3$ forces $3x^2 = 4^k-1$ and $y = -2^k$. Now, $3x^2 = (2^k-1)(2^k+1)$. It follows that one of $2^k-1$ or $2^k+1$ is a perfect square. By classics, this forces $k = 1$ or $k = 3$. Of these, $k = 3$ does not work as $63$ is not of the form $3x^2$. Hence $k = 1$, $x = \pm 1$ and, $y = -2$. It follows that $b = 11$ and then $a = 5$. This gives us the solution $(a, b, n) \equiv (5, 11, 2)$. Now, we get to the real deal, the case when $b$ is even. First, because $b$ is even, so is $a$ and we can rewrite \[2(a/2)^3 = (b/2)^2+2^{n-2}\]If, at this point we have $n = 2$, then we must solve $2a^3 = x^2+1$. Time to work in $\mathbb{Z}[i]$. Using similar ideas from above, we deduce that $b/2+i$ is either of the form $(1+i)(x+iy)^3$ or $(1-i)(x+iy)^3$ (because $2$ factors into the primes $1+i, 1-i$ in $\mathbb{Z}[i]$). This, after some easy observations yields only $b/2 = 1$ and $a/2 = 1$ as a valid solution, and we retrieve the solution $(a, b, n) \equiv (2, 2, 2)$. Henceforth $n > 2$ and onward we continue! Now if $n = 3$, then we have $(a/2)^3 = 1+2(b/4)^2$. Work in $\mathbb{Z}[\sqrt{-2}]$ to obtain that this has NO solutions with $a, b$ positive. Thus $n \geq 4$. We rewrite yet again \[4(a/4)^3 = (b/4)^2+2^{n-4}\]Fortunately, in this case because perfect squares are congruent to $0, 1$ modulo $4$ we promptly obtain that $4 \mid 2^{n-4}$ and we can finally reduce to $(a/4)^3 = (b/8)^2+2^{n-6}$. To sum up the argument we observed in this case, we obtain that any solution $(a, b, n)$ with $n$ even is either $(2, 2, 2)$ or obtained from a previous solution via $(a, b, n) \mapsto (4a, 8b, n+6)$. Thus, to grandly conclude, all solutions are given by $(a, b, n) \equiv (2^{2m} A, 2^{3m} B, C+6m)$ for some integer $m \geq 0$ and triple $(A, B, C) \in \{(2, 2, 2); (3, 5, 1); (5, 11, 2)\}$. $\square$