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.

The following argument has a lot of cases, but at least works (I think) to show the only solution is $2,1$.
Let $p(x,y)=4x^3+6x^2+3x+y+1.$ [I changed notation; $1 \le k \le l$ will be here $1 \le y \le x$], and then one wants each of $p(x,y)$ and $p(y,x)$ to be powers of $2$. Since the variables are each $\ge 1$ these powers of $2$ are at least $4+6+3+1+1=15,$ i.e. each is at least the fourth power of two.
Mod $2$, $p(x,y)=x+y+1$ so that the latter must be even, in particular $x\neq y$ so that $x>y.$ Now let $p(x,y)=2^b,\ p(y,x)=2^a$ and since $x>y$ implies $p(x,y)>p(y,x)$ we have $2^b>2^a$. So we have $b>a\ge 4.$ Our first step is to consider the sum $S=p(x,y)+p(y,x)$ which is $2^a+2^b=2^a(2^c+1)$ where $c=b-a\ge 1.$  we have
$$S=2(x+y+1)(2(x^2-xy+y^2)+x+y+1). \tag{1}$$
Here since $x+y+1$ is even it is coprime to $2^c+1$ and so divides $2^a$, making it a power of $2,$ say $2^r$. And here $r \ge 2$ since $x+y+1\ge 3.$ In this notation $S=2^{r+2}(x^2-xy+y^2+2^{r-1}).$ Since $2^{r-1}$ is still even, and $x^2-xy+y^2$ is odd because $x,y$ have opposite parity, we now see that the $2$-power in $S$, namely $2^a,$ must match the two power in the last expression for $S$, namely $2^{r+2}.$ This gives $r+2=a.$ 
We have arrived at $x+y+1=2^{a-2}.$  We apply this to $p(y,x)=2^a.$ We can write
$$p(y,x)=4y^3+6y^2+2y+(x+y+1)=4y^3+6y^2+2y+2^{a-2},$$
and so $g(y) \equiv 4y^3+6y^2+2y=2^a-2^{a-2}=3\cdot 2^{a-2}.$ This suggests looking at the congruence classes for $y$ mod $3$, dividing $g(y)$ through by $3$, and the result is to be  $2^{a-2},$ a power of $2$ which is at least $4$.
If $y=3t$ then $g/3=2t(18t^2+9t+1).$ Here $t=1$ leads to the second factor as $28$, so that cannot divide a power of $2$. Otherwise $t$ is even, making the second factor odd. So $y=3t$ is impossible.
If $y=3t-1$ then $g/3=2t(3t-1)(6t-1)$ has the odd factor $6t-1$ so $y=3t-1$ is not possible.
The remaining case is $y=3t+1$ in which $g/3=2(3t+1)(6t^2+7t+2).$ So in this case (because of the last factor $6t^2+7t+2,$) we need $t$ even, and if it is not zero then the factor $3t+1$ is odd and greater than $1$, not possible.
The conclusion so far of this is that it must be that we're in the $y=3t+1$ with $t=0$, that is $y=1$ is forced.
Now returning to $x+y+1=2^r$ where $r=a-2$ and filling in the known $y=1$ we have $x=2^r-2=2(2^{r-1}-1).$  The least $r$ here being $2$, this means that either $x=2$ or else $x$ is of the form $2(4t-1).$ However if the latter occurs then when $p(x,1)$ is computed one gets
$$p(2(4t-1),1)=4(512t^3-288t^2+54t-3),$$
which is not a power of $2$ since the factor in parentheses is odd.
Thus we have $x=2,y=1$ forced by the assumptions, as desired.
[Thanks to yoyostein (OP) for catching a sign difference on the $xy$ term of the quadratic involved in the factorization of $p(x,y)+p(y,x),$ which is now $x^2-xy+y^2$ as it should be.]
