The answer is $(2^k,2^k)$ where $k$ is a nonnegative integer. (solution) $(a+b)^3-2a^3-2b^3=3a^2b+3ab^2-a^3-b^3=3ab(a+b)-(a+b)(a^2-ab+b^2)=(a+b)(3ab-a^2+ab-b^2)=(a+b)(4ab-a^2-b^2)=2^t$ where $t$ is a positive integer. let the $gcd(a,b)=d$ and $a=dA$ and $b=dB$ $(a+b)(4ab-a^2-b^2)=d(A+B)d^2(4AB-A^2-B^2)=2^t$ from here we know $d^3|2^t$ so $d$ must be of the form $2^k$ where $k$ is a nonnegative integer. $(A+B)(4AB-A^2-B^2)=2^s$ $gcd(A+B,4AB-A^2-B^2)=gcd(A+B,2AB)|2$ (This is because $gcd(A,B)=1$) so $gcd(A+B,4AB-A^2-B^2)= 1$ or $2$ (Case 1) $gcd(A+B,4AB-A^2-B^2)=1$ this means either $A+B=1$ and $4AB-A^2-B^2=2^s$ or $A+B=2^s$ and $4AB-A^2-B^2=1$ both of them give contradiction $(mod 2)$ (Case 2)$gcd(A+B,4AB-A^2-B^2)=2$ here either (scenario 1)$A+B=2$ and $4AB-A^2-B^2=2^{s-1}$ or (scenario 2)$A+B=2^{s-1}$ and $4AB-A^2-B^2=2$ Lets look at scenario 1 $A+B=2$ and since $A,B$ are both positive integers they must be both 1. plugging this into $4AB-A^2-B^2$ we get 2 which means $s=2$. looking at scenario 2, $4AB-A^2-B^2=6AB -(A+B)^2 =6AB-2^{2(s-1)}=2$ looking at this $mod 3$ we get $2^{2(s-1)}=2 (mod 3)$ which is a contradiction. The only pair $(A,B)$ that works is $(1,1)$ and so the only $(a,b)$ that work is $(d,d)$ where $d$ is any number of the form $2^k$.

@above we have the additional solution $\left(2^k,3\times2^k\right)$. WLOG $a\leq b$. Since the equation is homogeneous, we can scale down $a,b$ until at least one is odd. Now $$(a+b)^3-2a^3-2b^3=(a+b)(4ab-a^2-b^2),$$so each of these factors is a power of $2$. It follows that $a,b$ are both odd. Hence $$4ab-a^2-b^2\equiv2\pmod4\implies4ab-a^2-b^2=2.$$Setting $u=\frac{a+b}{2}$ and $v=\frac{a-b}{2}$, this is equivalent to $u^2-3v^2=1$. This is a Pell equation; the solutions for $u$ are given by the recurrence $u_{n+1}=4u_n-u_{n-1}$, and we can check that $u_n\not\equiv0\pmod4$. Hence $u=1$, whence $a=b=1$, or $u=2$, whence $a=1,b=3$.