This answer is a bit late; sorry for that.
Kummer's proof of the nonsolvability of $x^p + y^p = z^p$ for regular primes $p$
used “ideal numbers” (in present-day language: ideals) and was intact, at least
basically. Hilbert in his Zahlbericht gave a modified proof. Both proofs cover not
only rational integers but also numbers in $\mathbb{Z}[\zeta_p]$. On the other
hand, Kummer’s second result concerning irregular primes that satisfy certain
additional conditions covers just the rational integers (although Hilbert, in the very
last section of Zahlbericht, erroneously says that Kummer had proven this result for
$\mathbb{Z}[\zeta_p]$ as well). Thus one cannot exclude the possibility that there
is indeed a solution $(x,y,z)$ for $p=37$. And because of "Kolyvagin's criterion"
about $(2^{37}-2)/37$, this solution must belong to the second case,
that is, at least one of these three numbers $x,y,z$ in $\mathbb{Z}[\zeta_{37}]$
must have a common factor with $37$ (as mentioned by George Lowther).
By the way, this criterion was also proven by Taro Morishima in 1935 (Japan. J.
Math. 11, 241-252, Satz 1; but warning: Satz 2 or at least its proof is incorrect since
it is based on some incorrect result of Vandiver).
I don’t know how to find such a solution $(x,y,z)$.
Here are a few partial results (the most interesting is probably (3)) towards this problem, but I don't have any hope that these will lead to a general solution. Your problem is a little similar to the famous Erdos-Moser question (which still remains unsolved): does there exist a solution to $1^n+2^n+\ldots+k^n= (k+1)^n$ apart from $1+2=3$?
$1$. If $k$ is even and there is a solution to the equation, then $k \equiv 2\pmod{16}$.
Proof: We may suppose that $k\ge 4$, and let $2^a$ exactly divide $k$. Then $x^k+ \ldots+(x+k-1)^k \equiv k/2 \pmod{2^{a+2}}$, while $(x+k)^k$ is either $0$ or $1 \pmod{2^{a+2}}$. It follows that $a=1$ and that $k/2\equiv 1\pmod 8$, or $k\equiv 2\pmod {16}$ as claimed.
$2$. If $k\equiv 1\pmod 4$ then the equation has no solutions. For $(x+1)^k+\ldots+(x+k-1)^k$ must be even, whereas $(x+k)^k-x^k$ is odd.
$3$. If $k\ge 5$ is odd and squarefree then the equation has no solutions.
Proof: Let $p$ be a prime dividing $k$. Note that $(p-1)\nmid k$. We make use of the fact that if $p$ is odd and $(p-1)\nmid k$ then $1^k+2^k+\ldots+(p-1)^k +p^k\equiv 0\pmod {p}$; this follows easily by picking a primitive root $\pmod p$ and summing the resulting geometric series. Therefore $x^k+\ldots+(x+k-1)^k \equiv 0\pmod p$, and so if this equals $(x+k)^k$ then $p$ must divide $x$. Since $k$ was assumed to be square-free, we conclude that $k|x$. But now note that $0^k+1^1+\ldots+(k-1)^k < k^k$ (for $k\ge 5$), and that
$(\ell k)^k+\ldots + (\ell k +k-1)^k > ((\ell+1)k)^k$ for all $\ell \ge 1$ and $k\ge 5$.
$4$. If $k$ is odd, and $p$ is an odd prime dividing $k-1$ such that $(k,p-1)=1$ then the equation has no solutions. In particular if $k \equiv 1\pmod{6}$, or $1\pmod {10}$ or $1\pmod{p(p-1)}$ for any odd prime $p$, there are no solutions.
Proof: If $p|(k-1)$ and $(p-1)\nmid k$ then we find from the observation used in (3) that
$(x+1)^k+\ldots+(x+k-1)^k \equiv 0 \pmod {p}$. Thus $(x+k)^k \equiv x^k \pmod p$. Since $(k,p-1)=1$ by assumption the map $x\to x^k \pmod {p}$ permutes all the residue classes $\pmod p$. Thus it follows that $(x+k)\equiv x \pmod p$, but this is impossible since $x+k \equiv x+1 \pmod p$.
Best Answer
This is still way open, I should think. "Elementary" methods won't even solve the analogous problem over $\mathbf{Z}$, so you need to use "modular form" methods. The problem is that even if the result were to follow from a Frey curve argument and a potential theorem of the form "all sufficiently nice Galois representations come from automorphic forms", we're a long way from establishing such a theorem.
A few more details: a key problem is that there are two natural candidates for where the automorphic forms will come from, and neither is good enough. The first is the group $GL(2)/\mathbf{Q}(i)$. This is a very natural place to look, but the problem is that this group does not admit Shimura varieties, so it's very hard to even go the "easy" way and to attach a Galois representation to an algebraic automorphic representation (i.e. do what Deligne did), let alone to do what Wiles did. There are some theorems of this nature, but they all have a self-duality hypothesis built into them, which will not hold for the Tate module of the Frey curve in general.
The second place to look is rank 2 unitary groups for the extension $\mathbf{Q}(i)/\mathbf{Q}$. These do have Shimura varieties and their geometry is understood fairly well nowadays, and there are are very strong theorems attaching Galois representations to these automorphic forms (there are even very strong theorems for rank $n$ unitary groups nowadays -- see for example the book by Harris and Taylor, but things have moved on even further since then, e.g. because of recent work of Sug-Woo Shin). However we then run into the same problem -- the unitary groups have some extra symmetry and this means that the associated Galois representations have some extra symmetry (they need to be essentially conjugate self-dual), and this extra symmetry will not in general be true for the Galois representations attached to the Tate module of the Frey curve.
So one needs a good new idea before we can push forward what one might call "Wiles' strategy" in this situation.
This is in marked contrast to the case of totally real fields, where a lot of the machinery works fine and it would not surprise me nowadays if FLT could be proved for several totally real fields. As has been implicitly mentioned in the comments, Jarvis and Meekin did this for $\mathbf{Q}(\sqrt{2})$, and this was a few years ago now, and modularity lifting theorems have moved on tremendously since then, so it would not surprise me if the experts could prove these results for other totally real number fields now. However somehow, after the Jarvis-Meekin work, which is a proof that the machine can be made to work in other cases, perhaps the interesting question now is not something like "is FLT true for $\mathbf{Q}(\sqrt{5})$?" (which one could perhaps hope to answer, perhaps with a lot of work, but with basically existing methods and a lot of hard graft to deal with the small $n$ cases) but more like "is some slightly weakened version of FLT true for all totally real fields?" or some such thing. You need to weaken it a bit because the case $n=3$ is an elliptic curve and it has positive rank for lots of totally real $F$, so now you need to deal with $n=4$ and $n=9$ and $n=6$ by hand, or just declare that you're only interested in $x^p+y^p=z^p$ with $p\geq5$ prime; and then you'll sometimes get reducible mod $p$ Galois representations -- so perhaps the correct weakening is something like "$x^p+y^p=z^p$ has no solutions for $p$ sufficiently large (depending on $F$)". Even then there may be problems in "case 2".
My impression is that the machinery being developed now is not really being developed with generalisations of FLT in mind, but it's generalising this "Wiles machine" in different directions, for example to prove things like the Fontaine-Mazur conjecture and the Sato-Tate conjecture. This is the direction deemed "trendy" -- and in a sense I can see why, because is FLT over a specific number field other than the rationals really a "natural" question? And FLT over a general number field is obviously false, so now you have to weaken things etc etc. On the other hand hard conjectures like Sato-Tate do look to me like natural questions.
IMPORTANT EDIT: I wrote this answer a long time ago -- what is it, 9 days now? -- but life moves on, and I hear from my spies in Toronto that Richard Taylor yesterday announced some results joint with Harris, Lan and Thorne, where they claim that they can attach Galois representations to (not necessarily self-dual) cohomological cuspidal automorphic representations on $GL(n)$ over totally real and CM fields. In particular apparently the theory for $GL(2)/\mathbf{Q}(i)$ is now up to about the state that the theory for $GL(2)/\mathbf{Q}$ was in the late 60s. So give it another 23 or so years and we should have FLT for $\mathbf{Q}(i)$!