A Non-Commutative Nullstellensatz In studying presentations of pro-$p$-groups via generators and relations, one is led (via the so-called Magnus embedding) to questions involving power series in non-commuting variables.  Results from local algebraic geometry occasionally shed some insight on how to make progress, but more often that not, I find myself lacking appropriate analogs of major theorems from the commutative case.  I haven't had much luck in books on non-commutative ring theory or non-commutative algebraic geometry -- the focus seems to be on completely different ideas (though I'll happily stand corrected).  In any case, here's an important and seemingly basic question that I don't know how to answer.

Let $\mathbb{F}_p\langle\langle x,y\rangle\rangle$ be the ring of formal power series over $\mathbb{F}_p$ in two non-commuting variables $x$ and $y$.  This ring has a unique two-sided maximal ideal $I=(x,y)$.  Suppose $f,g\in I$.  Can anything be said about the smallest $n$, if one exists, such that $I^n\subset (f,g)$?  Namely, when does this quantity exist?  Is this quantity computable?  Boundable?

It's trivial to come up with examples for which there is no $n$, e.g., $(xy,yx)$, since no $x^n$ is contained in this ideal.  I'm not sure how exactly to quantify this observation.  Is there some kind of non-commutative resultant at play here?
Edit:  I think it might be helpful for me to update with some examples as we go along.  Here's one that I thing captures at least some of the interesting parts of this question.
Take $p=3$, $f=x+y$, and $g=x^3$.  Then the inclusion $I^3\subset (f,g)$ can be seen by taking each of the 8 monomials in $I^3$ verifying that they are in $(f,g)$, e.g., $yxy=yfy-f^3+g\in (f,g)$.  The same argument applies with the same $f$ and taking $g=x+y+x^3$.  This seems to me evidence that this question can't be answered only by looking at the leading monomials (though admittedly it might be easy enough to exclude these trivial counter-examples).  
 A: Suppose that $f$ and $g$ are monomials and that $(f,g)$ contains a power of $I$. Then every word of sufficient length must contain $f$ or $g$ as a subword (and conversely). Thus your argument with $x^n$ shows that either $f$ or $g$ is a power of $x$, and likewise, one of them is a power of $y$. If $f=x^k, g=y^m, k,m\geq 2$ then the word $(x^{k-1}y)^N$ can be arbitrarily long and doesn't contain $f$ or $g$ — contradiction. Thus up to relabelling, $f=x^k, g=y$ and $n=k$ is minimal with the property that $I^n\subset (f,g)$.  
I recommend Algebraic Combinatorics on Words by M. Lothaire (google it) for related ideas.
A: Let $F$ be a field, and let $f_1,f_2,\ldots, f_k\in R:=F\langle\langle x,y\rangle\rangle$ with $k\in \mathbb{N}$.  Order monomials in $R$ by degree, and then lexicographically.  Since the question concerns computability, assume that there is an algorithm which spits out the coefficients of the monomials in $f_i$ (in order).  I claim that:
Proposition: Given any fixed $n\in \mathbb{N}$, there is an algorithm to decide whether $I^n\subseteq (f_1,f_2,\ldots, f_k)$.
To prove this, first we need a lemma and some notation.  Given $r\in R$ we write $r[n]$ for the homogeneous component of $r$ in degree $n$.
Lemma: $I^n\subseteq (f_1,f_2,\ldots, f_k)$ iff for each of the $2^n$ monomials $m$ of total degree $n$ there is a power series $g_m\in (f_1,f_2,\ldots, f_k)$ such that $g_m[n]=m$.
Proof of the lemma: The forward direction is obvious.  For the reverse, let $m_1,m_2,\ldots, m_{2^n}$ be the list of all monomials of degree $n$, and let $g_i:=g_{m_i}$.  Fix $h\in I^n$.  We can write $h[n]=\sum m_i a_{i,n}$ for some $a_{i,n}\in R$ (all of degree $0$).  Setting $h':=h-\sum g_i a_{i,n}$, we see that $h'$ has zero homogeneous components in degree $\leq n$.  We can write $h'[n+1]=\sum m_i a_{i,n+1}$ for some $a_{i,n+1}\in R$ (all of degree $1$).  Set $h''=h-\sum g_i(a_{i,n}+a_{i,n+1})$.  Repeating this process, we obtain power series $a_i=\sum_{m\geq n}a_{i,m}$ such that $h=\sum_i g_i a_i\in (f_1,f_2,\ldots, f_k)$.$\qquad \blacksquare$
Proof of the proposition:  By the lemma, it suffices to decide for each monomial $m\in I^n$, whether there is a power series $g_m\in (f_1,f_2,\ldots, f_k)$ with $g_m[n]=m$.  We may as well work in the quotient ring $R/I^{n+1}$.  The image of the ideal $(f_1,f_2,\ldots, f_k)$ modulo $I^{n+1}$ is a finite-dimensional $F$-vector space, and so the question is easily decided by a row-reduction style argument.  [More concretely, one can modify the set $\{f_1,f_2,\ldots, f_k\}$ modulo $I^{n+1}$ so that no leading terms are linear combinations of others, and the set is closed under left and right multiplication by $x$ and $y$.  Then, the leading terms of degree $n$ either do or do not generate all of the appropriate monomials.]$\qquad \blacksquare$

To give an interesting example, consider $f_1=yx-y^2+x^3$, $f_2=xy-x^2$.  One can show directly (using the methodology above) that $I^{5}\subseteq (f_1,f_2)$.  If we modify $f_1=yx-y^2+x^k$ (with $k\geq 3$), then we get $I^{k+2}\subseteq (f_1,f_2)$.

On the question of whether it is decidable, given as input only the algorithms describing the power series $f_1,\ldots, f_k$, whether an $n$ exists for which $I^n\subseteq (f_1,f_2,\ldots, f_k)$, I don't know the answer.
