$3^k$ not congruent to $-1 \pmod {2^e}, e > 2$. $3^k \not\equiv -1 \pmod {2^e}$ for $e > 2, k > 0$. Is this true? I have tried to prove it by expanding $(1 + 2)^k$. [Notation: $(n; m) := n! / (m! (n - m)!)$] E.g., for $e = 3$ I get: $(1+2)^k + 1 = 2 + (k; 1) 2 + (k; 2) 2^2 + (k; e) 2^e + ...$ So, here it's enough to prove that $2^3$ does not divide $2 + (k; 1) 2 + (k; 2) 2^2$. The validity for general e seems very hard to prove. 

Just a bit more explanative in case the other answers are too compressed.
We do this by induction. The remarkable property, that $\small 3^2=9=8+1$ suggests, to try this modulo 8 (but that could as well be suggested by the problem and it was asked that the factor $\small 2^e $ with $\small e=2 $ should not be exceeded)      


*

*we observe for the first even exponent
$\small   3^2+1 = 10 = (8 \cdot 1   + 1) +1 $      

*we try the general ansatz
$\small   3^k+1 = (8  m_0 + 1) +1 $
then
$\small \begin{eqnarray}
3^{k+2}+1 &=& \\
9 \cdot 3^k+1 &=& (8+1)(8 m_0 + 1) +1 \\ 
&=& 8 (8 m_0+1) + (8 m_0 + 1) +1 \\ 
&=& 8 (9 m_0+1) +  1 +1 \\ &=& (8 m_1+1)+1  \end{eqnarray}$
and see, that the structure persists when we increase the exponent by 2. Because that structure is true for exponent k+2 when it is true for k, and we know, that it is true for k=2, we know it is true for all even exponents.

*Then we look at the odd exponents and observe for the first one
$\small 3^1+1 = 4 = (8 \cdot 0+3)+1 $       

*we do the general ansatz
$\small 3^k+1 = (8m+3)+1 $
and then
$\small \begin{eqnarray}
 9 \cdot 3^k+1 &=&  (8+1)(8m_0+3)+1  \\
   &=& 8(8m_0+3)+1 \cdot 8m_0+1 \cdot 3+1 \\ 
   &=& 8(9m_0+3)+4 \\
   &=& (8m_1+3)+1 \end{eqnarray} $
In 2) and 4) we conclude, that we have for even exponents k $\small 3^k \equiv 1 \pmod 8$ and for odd exponents $\small 3^k \equiv 3 \pmod 8$ and thus $\small 3^k \equiv -1 \pmod 4$ or  $\small 3^k \equiv -1 \pmod 2$ but not $\small \pmod 8$
