If $4k^3+6k^2+3k+l+1$ and $4l^3+6l^2+3l+k+1$ are powers of two, how to conclude $k=1, l=2$ It is given that 
$$4k^3+6k^2+3k+l+1=2^m$$
and 
$$4l^3+6l^2+3l+k+1=2^n$$
where $k,l$ are integers such that $1\leq k\leq l$.
How do we conclude that the only solution is $k=1$, $l=2$?
I tried subtracting the two equations to get:
$$2(l-k)(2l^2+2lk+2k^2+3l+3k+1)=2^m(2^{n-m}-1)$$
But I am unable to proceed further.
Thanks for any help!
Updates:
An observation is that $k$ and $l$ have opposite parity.
Because if both $k$ and $l$ are even (or both odd), then $4k^3+6k^2+3k+l+1\equiv 1 \pmod 2$, which is a contradiction.

I have a simplification that I couldn't finish, but maybe someone better with polynomials can. It may also lead nowhere, obviously; but was too long as a comment.    
If you add the two equations, you get
$$4(k^3 + l^3)+ 6(k^2 + l^2) + 4 (k + l) + 2 = 2^m + 2^n$$
If you look at this modulo (k+l), it implies (mod k+l):
$$4(k+l)^{3} + 6(k+l)^{2} + 2 = 2^m + 2^n$$
Substituting $x:= k+l$, we get 
$$4x^{3} + 6x^{2} + 2 = 2^m + 2^n$$
That's where I get stuck. 
