I don't have anything close to a solution, but a simple C++ script reveals the solutions to the first $10$ powers of $2$. Denoting $P(x, y) = 5xy - x^2 - 2y^2$, it can be computed that $$P(1, 1) = 2^1, P(3, 1) = 2^2, P(13, 3) = 2^3, P(3, 5) = 2^4, P(5, 3) = 2^5$$, and , $$P(19, 5) = 2^6, P(13, 27) = 2^7, P(11, 13) = 2^8, P(29, 11) = 2^9, P(67, 149) = 2^{10}$$. There does not seem to be any apparent pattern, and the odd number condition makes it so much harder!

Maybe can be solved by induction, finding this procedure: $5(ax+by)(cx+dy)-{{(ax+by)}^{2}}-2{{(cx+dy)}^{2}}={{2}^{k}}(5xy-{{x}^{2}}-2{{y}^{2}})$ for any natural $k$, and $a,b,c,d$ natural numbers.

I don't have a full solution either, but we know that $5xy-x^2-2y^2=2^k.$ We can write this is a quadratic in $x$ and $y$ and use the quadratic formula to find that $x = \dfrac{5y\pm\sqrt{17y^2-2^{k+2}}}{2}$ and $y =\dfrac{5x\pm\sqrt{17x^2-2^{k+3}}}{4}.$ Since $x, y$ are odd integers, the first expression is integral if and only if $17y^2-2^{k+2}$ is a perfect square. For the second expression, we can deduce its integral if and only if the expression in the radical is integral (although this is slightly harder). So, we have reduced the problem to proving that for all positive integer values of $k$ there exists integral $m, n$ such that $17(2n+1)^2-2^{k+3}=m^2.$ (We can plug in $k' = k+1$ to prove the other part.)

Let $\alpha:={{\sqrt{17}-5}\over 2}, R={\bf Z}[\alpha]$ then the norm function $N:R\rightarrow {\bf Z}$ is given by $N(x+y\alpha)=(x+y\alpha)(x+y{\overline \alpha})=x^2+xy(-5)+2y^2=x^2-5xy+2y^2$. If you don't know the norm function, the important property is that $N(ab)=N(a)N(b)$. Anyway, with $p:=1+\alpha,q:=1+2\alpha$ we have $Np=-2, Nq=-1$. Hence $Np=-2,N(qp^2)=-4,N(p^3)=-8,\cdots$ and it is not hard to show that all these elts have the form $r+s\alpha$ where $r,s$ are odd. Indeed, if $r,s$ are odd, then $(r+s\alpha)p^2=(r+s\alpha)(-3\alpha-1)=6s-r+(14s-3r)\alpha$ where $6s-r,14s-3r$ are odd.

TuZo wrote: Maybe can be solved by induction, finding this procedure: $5(ax+by)(cx+dy)-{{(ax+by)}^{2}}-2{{(cx+dy)}^{2}}={{2}^{k}}(5xy-{{x}^{2}}-2{{y}^{2}})$ for any natural $k$, and $a,b,c,d$ natural numbers. mathwiz0803 wrote: I don't have a full solution either, but we know that $5xy-x^2-2y^2=2^k.$ We can write this is a quadratic in $x$ and $y$ and use the quadratic formula to find that $x = \dfrac{5y\pm\sqrt{17y^2-2^{k+2}}}{2}$ and $y =\dfrac{5x\pm\sqrt{17x^2-2^{k+3}}}{4}.$ Since $x, y$ are odd integers, the first expression is integral if and only if $17y^2-2^{k+2}$ is a perfect square. For the second expression, we can deduce its integral if and only if the expression in the radical is integral (although this is slightly harder). So, we have reduced the problem to proving that for all positive integer values of $k$ there exists integral $m, n$ such that $17(2n+1)^2-2^{k+3}=m^2.$ (We can plug in $k' = k+1$ to prove the other part.) $k$ is the problem here

Define the recurrences $x_k = 5x_{k-1} - 2y_{k-1}$ and $y_k = x_{k-1}$, with $x_1 = 1$, $x_2 = 3$ and $y_1 = y_2 = 1$. Let's prove by induction that $5x_ky_k - x_k^2 -2y_k^2 = 2^k$ for every positive integer $k$. Cases $k = 1$ and $k = 2$ are trivial. Now suppose that $5x_ky_k - x_k^2 - 2y_k^2 = 2^k$. Then we have $5x_{k+1}y_{k+1} - x_{k+1}^2 - 2y_{k+1}^2$ $=$ $5(5x_k - 2y_k)x_k - (5x_k - 2y_k)^2 - 2x_k^2$ $=$ $10x_ky_k - 2x_k^2 - 4y_k^2$ $=$ $2(5x_ky_k - x_k^2 - 2y_k^2)$ $=$ $2\cdot2^k$ $=$ $2^{k+1}$, as we wanted to prove. Since $x_k$ and $y_k$ are always odd positive integers, we finish the problem.