Can anything be proven about this complex variant of the Collatz problem, or is it just as intractable?

The Collatz conjecture works heuristically because the map is contracting on average (to prove this it is easier to use $x \mapsto (3x+1)/2$ when $x$ is odd)

For your procedure this is no longer the case because dividing by $1+i$ divides the modulus by $\sqrt 2$.

If $z$ is "odd" then you multiply it by approximately $3/ \sqrt 2$ in two steps, and if it is "even" you divide it by $\sqrt 2$ in one step, so on average, you multiply a number by $3/2$ in $3$ steps, which means a multiplication by $(3/2)^{1/3}$ per step. This is greater than $1$ so heuristically, most of the time the sequence should diverge to infinity.

For example if you start at $1+2i$, after 200 steps you are at $-673097903812-335968886130i$, whose norm is approximately $7.523 \times 10^{11}$ and it doesn't look like it wants to get back to small numbers.

For comparison with the heuristic, $\sqrt 5 \times (3/2)^{200/3} \approx 1.227 \times 10^ {12}$, which is decently close.

The hope here is apparently to find that all Gaussian integers eventually lead to a number of the form $(1 + i)^n u$, where $n$ is a positive real integer and $u$ is a Gaussian unit.

Looking at a list of powers of $1 + i$, it seems to me like they take one of these forms: $\pm 2^n$, $\pm 2^n i$, $\pm 2^n \pm 2^n i$.

Then it's definitely possible for, say, $-3i$ to be followed by $-8i$ and form there it's a spiral drop to 1.

But when the real and imaginary parts are both nonzero, we run into the problem that $3z + i$ affects the real and imaginary parts unevenly.

For example, $5 + 5i$ multiplied by 3 is $15 + 15i$, but then you're only adding $i$ and so $15 + 16i$ falls short of $16 + 16i$ by 1.

Since $\mathbb Z[i]$ is a unique factorization domain and 3 is prime, a number with real and imaginary parts both nonzero is divisible by 3 only if each part is itself also divisible by 3.

Hence $3z$ fails to lead us to a number of the form $2^n + (2^n - 1)i$. A similar problem occurs with $3z + 1$ instead of $3z + i$.

Not a full answer, just some remarks and considerations.

In the norm language we can see that division of $z$ with $1+i$ produces some complex number $z^*$ that has the norm smaller by a factor of $\sqrt{2}$, and $3z+1$ approximately triples the norm.

Now since ${(\sqrt{2}})^3 \approx 2.8<3$ we see that for every $z \in \mathbb Z$ his associated Collatzian sequence must be generated in such a way that, on average, on every appliance of a function $f(z)=3z+1$ there are more than three appliances of a function $f(z)=\frac {z}{1+i}$ if we want that all subsequences of a Collatzian sequence associated to $z$ do not escape to infinity.

But this seems very unlikely to ever happen because there seems to be no reason that division $\frac {z}{1+i}$ will favour numbers of the form $O+Oi$ and $E+Ei$ quite more often than numbers $E+Oi$ and $O+Ei$, where $E$ stands for even and $O$ for odd integers.