Solve $x^2+x+7 \equiv 0\pmod{81}$ Solve $x^2+x+7\equiv 0 \pmod{81}$

My work:
Prime factorization $81 = 9^2 = 3^4$
Test the value $x\equiv0,1,2$ for $x^2+x+7\equiv0\mod{3}$
we have $x\equiv1\mod{3}$ works.
Now life this to $\mod{3^2} = \mod{9}$
Let $x=1+3k$ for some integer$k$
$(1+3k)^2 + (1+3k) +7 \equiv0\mod{9}$
$1+6k+0+1+3k+7\equiv0\mod{9}$
$9+9k\equiv0\mod9$
$1+k\equiv0\mod9$
$k\equiv-1\mod9$
$k=-1+9m$ for some integer m
Lift again to $\mod81$
$(-1+9m)^2 + (-1+9m)+7\equiv0\mod81$
$-9m+7\equiv0\mod81$
I cant continue... how can I make it work???
thanks!!
 A: You can write the equation as $(x +41)^{2} \equiv -27$ (mod 81). But this means there is no solution, as $x+41$ would need to be divisible by $9.$ I'll put in more explanation, in view of comments and questions, including correcting a sign error pointed out in the comments.
We have $x^{2} + x + 7 \equiv x^{2}+ 82x + 7 $ (mod 81). This is (mod 81), equal to $(x+41)^{2}+ 7 -1600,$ and also $1600 \equiv -20$ (mod 81). Hence we need 
$(x+41)^{2} \equiv -27$ (mod 81), which is impossible in integers.
A: Let's argue following your approach: As you say, $x$ must be of the form $3k+1$. We then have that $$x^2+x+7=9k^2+9k+9=9(k^2+k+1).$$ This is a multiple of $81=9\times 9$ iff $9$ divides $k^2+k+1$. Again, arguing as you did, we see that $k$ must have the form $3t+1$ for some integer $t$, and then $$k^2+k+1=9t^2+9t+3=9(t^2+t)+3,$$ which never is a multiple of $9$, so there are no solutions.
A: Let us do the lifting, patiently. As was pointed out, the solution $x\equiv 1\pmod{3}$ lifts to three solutions modulo $3^2$, namely $1$, $4$, and $7$.
Let us try to now lift to $3^3$. We do only the case $x\equiv 1\pmod 3^2$. Let $x=1+9k$ for some $k$. Then $x^2+x+7\equiv 1+18k+1+9k+7\equiv 27k+9\pmod{3^3}$. It is clear that there is no $k$ that works. A similar calculation shows that the solutions $x\equiv 4\pmod{3^2}$ and $x\equiv 7\pmod{3^2}$ do not lift to $3^3$. 
