As we saw on this page the nature of powers depends on the nature of the square.
Square
m = a + b e_{1} + c e_{2} + d e_{12}
m² = (a + b e_{1} + c e_{2} + d e_{12})(a + b e_{1} + c e_{2} + d e_{12})
multiplying out the terms gives:
m² = 

cancelling out the anticommuting terms gives:
m² =a²+b²+c²d² + 2(a*b*e_{1}+a*c*e_{2}+a*d*e_{12})
To simplify this a bit let s= b²+c²d²a² so we get:
m² =s + 2a²+ 2(a*b*e_{1}+a*c*e_{2}+a*d*e_{12})
m² =s + 2a*m
Higher Order Powers
Using the above expression for the square we can generate all other powers from the term before it since:
if m^{n}=u + v*m
where u and v are arbitary scalars then:
m^{n+1}=u*m + v*m²
m^{n+1}=u*m + v*(s + 2a*m)
m^{n+1}=v*s + (u + v*2a)*m
so to derive the next term we use:
u' = v*s
v' = u + v*2a
which gives:
n  u  v 

1  0  1 
2  s  2a 
3  2a*s  4a²+s 
4  s² + 4*a²*s  4*s*a + 8*a³ 
5  4*s²*a + 8*s*a³  s² + 12*a²*s + 16*a^{4} 
Can we deal with the commuting terms and the anticommuting terms seperately? The commuting terms lead to binomial(or trinomial?) elements and the anticommuting terms lead to elements where even powers are scalar and odd powers are scalar multipiers of the original multivector.
Can we find such a pattern? Lets try the next term:
m³ =(s + 2a*m)*m
m³ =s*m + 2a*m²
m³ =s*m + 2a*(s + 2a*m)
m³ =s*m + 2a*s + 4a²*m
m³ =2a*s + (4a²+s)*m
So it seems that all powers seem to be a linear function of m
next m^{4}
m^{4} = m²*m²
m^{4} = (s + 2a*m)*(s + 2a*m)
m^{4} = s² + 4*s*a*m + 4*a²*m²
m^{4} = s² + 4*s*a*m + 4*a²*(s + 2a*m)
m^{4} = (s² + 4*a²*s) + (4*s*a + 8*a³)*m
next m^{5}
m^{5} = (s² + 4*a²*s)*m + (4*s*a + 8*a³)*m²
m^{5} = (s² + 4*a²*s)*m + (4*s*a + 8*a³)*(s + 2a*m)
m^{5} = (4*s²*a + 8*s*a³) + (s² + 12*a²*s + 16*a^{4})*m