I did the following experiment:
Let $p$ be a prime number. Then a necessary condition for the sequence to remain in $\mathbb{Z}$ is that $x_{p-1} \equiv \pm 1 \mod p$.
So for every starting value $x_1$, I calculated $x_{p-1} \mod p$ for the first several prime numbers $p$ to see if there are obstructions. It turns out that for every choice of $x_1$ between $2$ and $100000$, there are always obstructions. The first obstruction (i.e. the smallest prime $p$ such that $x_{p-1}$ is not congruent to $\pm 1$ modulo $p$) for $2 \leq x_1 \leq 100$ is listed below.
x[1] = 2: obstruction at 2
x[1] = 3: obstruction at 5
x[1] = 4: obstruction at 2
x[1] = 5: obstruction at 5
x[1] = 6: obstruction at 2
x[1] = 7: obstruction at 5
x[1] = 8: obstruction at 2
x[1] = 9: obstruction at 23
x[1] = 10: obstruction at 2
x[1] = 11: obstruction at 7
x[1] = 12: obstruction at 2
x[1] = 13: obstruction at 5
x[1] = 14: obstruction at 2
x[1] = 15: obstruction at 5
x[1] = 16: obstruction at 2
x[1] = 17: obstruction at 5
x[1] = 18: obstruction at 2
x[1] = 19: obstruction at 11
x[1] = 20: obstruction at 2
x[1] = 21: obstruction at 7
x[1] = 22: obstruction at 2
x[1] = 23: obstruction at 5
x[1] = 24: obstruction at 2
x[1] = 25: obstruction at 5
x[1] = 26: obstruction at 2
x[1] = 27: obstruction at 5
x[1] = 28: obstruction at 2
x[1] = 29: obstruction at 13
x[1] = 30: obstruction at 2
x[1] = 31: obstruction at 7
x[1] = 32: obstruction at 2
x[1] = 33: obstruction at 5
x[1] = 34: obstruction at 2
x[1] = 35: obstruction at 5
x[1] = 36: obstruction at 2
x[1] = 37: obstruction at 5
x[1] = 38: obstruction at 2
x[1] = 39: obstruction at 7
x[1] = 40: obstruction at 2
x[1] = 41: obstruction at 11
x[1] = 42: obstruction at 2
x[1] = 43: obstruction at 5
x[1] = 44: obstruction at 2
x[1] = 45: obstruction at 5
x[1] = 46: obstruction at 2
x[1] = 47: obstruction at 5
x[1] = 48: obstruction at 2
x[1] = 49: obstruction at 7
x[1] = 50: obstruction at 2
x[1] = 51: obstruction at 19
x[1] = 52: obstruction at 2
x[1] = 53: obstruction at 5
x[1] = 54: obstruction at 2
x[1] = 55: obstruction at 5
x[1] = 56: obstruction at 2
x[1] = 57: obstruction at 5
x[1] = 58: obstruction at 2
x[1] = 59: obstruction at 7
x[1] = 60: obstruction at 2
x[1] = 61: obstruction at 11
x[1] = 62: obstruction at 2
x[1] = 63: obstruction at 5
x[1] = 64: obstruction at 2
x[1] = 65: obstruction at 5
x[1] = 66: obstruction at 2
x[1] = 67: obstruction at 5
x[1] = 68: obstruction at 2
x[1] = 69: obstruction at 11
x[1] = 70: obstruction at 2
x[1] = 71: obstruction at 11
x[1] = 72: obstruction at 2
x[1] = 73: obstruction at 5
x[1] = 74: obstruction at 2
x[1] = 75: obstruction at 5
x[1] = 76: obstruction at 2
x[1] = 77: obstruction at 5
x[1] = 78: obstruction at 2
x[1] = 79: obstruction at 29
x[1] = 80: obstruction at 2
x[1] = 81: obstruction at 7
x[1] = 82: obstruction at 2
x[1] = 83: obstruction at 5
x[1] = 84: obstruction at 2
x[1] = 85: obstruction at 5
x[1] = 86: obstruction at 2
x[1] = 87: obstruction at 5
x[1] = 88: obstruction at 2
x[1] = 89: obstruction at 13
x[1] = 90: obstruction at 2
x[1] = 91: obstruction at 7
x[1] = 92: obstruction at 2
x[1] = 93: obstruction at 5
x[1] = 94: obstruction at 2
x[1] = 95: obstruction at 5
x[1] = 96: obstruction at 2
x[1] = 97: obstruction at 5
x[1] = 98: obstruction at 2
x[1] = 99: obstruction at 11
x[1] = 100: obstruction at 2
Up to $x_1 = 100000$, the biggest "first obstruction" appears at:
x[1] = 13589: obstruction at 103
Even if one allows $x_1$ to be $\sqrt{k}$ for some integer $k$, the results are similar - one just starts from $x_2$, and for $2 \leq x_2 \leq 10000$ there are always obstructions at (small) prime numbers.
These results seem to support the original conjecture.
EDIT
Following this idea, I further calculated, for a given prime $p$, the residue classes of $x_1 \mod p$ that will lead to an obstruction at $p$. Let us call them "bad" residues. The result seems to be interesting for its own sake.
p bad residues x[1] mod p
mod 2: 0
mod 3:
mod 5: 0 2 3
mod 7: 0 3 4
mod 11: 0 3 5 6 8
mod 13: 0 2 3 5 8 10 11
mod 17: 2 4 5 12 13 15
mod 19: 0 6 8 11 13
mod 23: 3 7 8 9 11 12 14 15 16 20
mod 29: 0 3 5 6 7 8 10 11 13 14 15 16 18 19 21 22 23 24 26
mod 31: 0 2 5 8 10 13 15 16 18 21 23 26 29
mod 37:
mod 41:
mod 43: 0 2 3 4 5 6 7 8 9 10 11 14 16 18 19 20 21 22 23 24 25 27 29 32 33 34 35 36 37 38 39 40 41
mod 47: 0 7 8 9 10 11 12 13 14 15 18 21 23 24 26 29 32 33 34 35 36 37 38 39 40
mod 53: 0 9 12 15 17 18 20 23 30 33 35 36 38 41 44
mod 59: 3 5 8 13 14 15 16 18 21 26 33 38 41 43 44 45 46 51 54 56
mod 61: 0 2 5 8 11 12 15 17 19 20 21 22 23 24 26 29 32 35 37 38 39 40 41 42 44 46 49 50 53 56 59
mod 67:
mod 71: 5 6 15 19 20 24 25 31 33 35 36 38 40 46 47 51 52 56 65 66
mod 73: 9 23 27 28 29 44 45 46 50 64
mod 79:
mod 83:
mod 89: 0 2 3 4 5 16 17 22 23 24 27 30 31 32 35 40 49 54 57 58 59 62 65 66 67 72 73 84 85 86 87
mod 97: 0 2 3 8 11 14 15 17 21 23 24 28 29 30 35 38 39 44 47 50 53 58 59 62 67 68 69 73 74 76 80 82 83 86 89 94 95
And here is the table which counts the number of bad residues modulo $p$:
p number of bad residues x[1] mod p
mod 2: 1
mod 3: 0
mod 5: 3
mod 7: 3
mod 11: 5
mod 13: 7
mod 17: 6
mod 19: 5
mod 23: 10
mod 29: 19
mod 31: 13
mod 37: 0
mod 41: 0
mod 43: 33
mod 47: 25
mod 53: 15
mod 59: 20
mod 61: 31
mod 67: 0
mod 71: 20
mod 73: 10
mod 79: 0
mod 83: 0
mod 89: 31
mod 97: 37
mod 101: 50
mod 103: 35
mod 107: 29
mod 109: 20
mod 113: 30
mod 127: 22
mod 131: 93
mod 137: 33
mod 139: 115
mod 149: 121
mod 151: 59
mod 157: 6
mod 163: 111
mod 167: 85
mod 173: 111
mod 179: 98
mod 181: 127
mod 191: 0
mod 193: 83
mod 197: 4
mod 199: 130
mod 211: 85
mod 223: 34
mod 227: 77
mod 229: 57
mod 233: 85
mod 239: 137
mod 241: 56
mod 251: 140
mod 257: 79
mod 263: 0
mod 269: 44
mod 271: 129
mod 277: 20
mod 281: 26
mod 283: 231
mod 293: 171
The most noticeable thing, to me, is those primes with "$0$" bad residues. Here are they:
3, 37, 41, 67, 79, 83, 191, 263, 347, 353, 373, 379, 421, 449, 463, 509, 557, 619, 647, 661, 673, 719, 733, 757, 787, 823, 839, 911
This sequence is not found in OEIS.
Let us call those primes "exceptional". If one excludes those exceptional primes, then the proportion of bad residues (i.e. [number of bad residues mod $p$] divided by $p$) seems to distribute uniformly on the interval $[0, 1)$. This suggests that the exceptional primes may of particular interest.
EDIT
To illustrate the distribution of the proportion of bad residues, here I add the statistical data (for primes $p < 5000$):
"bad proportion" number of primes
0 77 (i.e. number of exceptional primes)
(0.00, 0.02] 10
(0.02, 0.04] 15
(0.04, 0.06] 13
(0.06, 0.08] 18
(0.08, 0.10] 7
(0.10, 0.12] 11
(0.12, 0.14] 8
(0.14, 0.16] 14
(0.16, 0.18] 21
(0.18, 0.20] 14
(0.20, 0.22] 16
(0.22, 0.24] 11
(0.24, 0.26] 17
(0.26, 0.28] 15
(0.28, 0.30] 13
(0.30, 0.32] 11
(0.32, 0.34] 15
(0.34, 0.36] 15
(0.36, 0.38] 17
(0.38, 0.40] 17
(0.40, 0.42] 19
(0.42, 0.44] 13
(0.44, 0.46] 15
(0.46, 0.48] 23
(0.48, 0.50] 20
(0.50, 0.52] 16
(0.52, 0.54] 11
(0.54, 0.56] 15
(0.56, 0.58] 16
(0.58, 0.60] 14
(0.60, 0.62] 12
(0.62, 0.64] 6
(0.64, 0.66] 13
(0.66, 0.68] 20
(0.68, 0.70] 14
(0.70, 0.72] 8
(0.72, 0.74] 9
(0.74, 0.76] 9
(0.76, 0.78] 6
(0.78, 0.80] 11
(0.80, 0.82] 12
(0.82, 0.84] 7
(0.84, 0.86] 6
(0.86, 0.88] 5
(0.88, 0.90] 6
(0.90, 0.92] 3
(0.92, 0.94] 3
(0.94, 0.96] 2
(0.96, 0.98] 0
(0.98, 1.00] 0
There is clearly a concentration on $0$, i.e. on the exceptional primes.
The "average proportion", calculated as $\frac{\sum_p proportion_p}{\sum_p 1}$, is about $0.37551$.
Best Answer
As remarked by Joe Silverman, a natural way to look at this question is by phrasing it in terms of the map $f(x):=\frac{1}{2}x(x+1)$ on the $2$-adic integers $\mathbb{Z}_2$. We are then asking about the behaviour of the orbit $f^n(2)$ with respect to the partition of $\mathbb{Z}_2$ into two clopen sets $U_1:=2\mathbb{Z}_2$ and $U_2:=1+2\mathbb{Z}_2$. Most questions of this form are very hard, for reasons which I will attempt to explain. Short answer: dynamically, there is no obvious reason why this should be easier than the Collatz problem or the normality of $\sqrt{2}$.
Suppose that we are given a continuous transformation $f$ of a compact metric space $X$ and a partition of $X$ into finitely many sets $U_1,\ldots,U_N$ of reasonable regularity (e.g. such that the boundary of each $U_i$ has empty interior). We are given a special point $x_0$ and we want to know how often, and in what manner, the sequence $(f^n(x_0))_{n=1}^\infty$ visits each of the sets $U_i$. We might believe the following result to be useful: if $\mu$ is a Borel probability measure on $X$ such that $\mu(f^{-1}A)=\mu(A)$ for every Borel measurable set $A\subset X$, and if additionally every $f$-invariant measurable set $A\subseteq X$ satisfies $\mu(A)\in \{0,1\}$, then $$\lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1}\phi(f^i(x)) =\int \phi\,d\mu$$ for $\mu$-almost-every initial point $x$ and every $\mu$-integrable function $\phi \colon X \to \mathbb{R}$. This is the Birkhoff ergodic theorem or pointwise ergodic theorem. In particular we could take $\phi$ to be the characteristic function of a partition element $U_i$, so the above measures the average time spent in $U_i$ by the sequence $f^n(x)$. Or we could take $\phi$ to be the characteristic function of a set such as $U_1\cap f^{-1}U_2 \cap f^{-2}U_1$, which would tell us how often the orbit of $x$ follows the path $U_1 \to U_2 \to U_1$, and so on.
The Birkhoff theorem is great if we are content to study points $x$ which are generic with respect to a particular invariant measure. If we want to study a specific $x$ then we have no way to apply the theorem. If multiple different invariant measures $\mu$ exist then the integrals which they assign to the characterstic functions $\chi_{U_i}$ will in general be different and we will get different answers, and the Birkhoff averages along orbits will have fundamentally different behaviours depending on which measure the starting point is generic with respect to (if any). In a broad sense this is an aspect of the phenomenon called "chaos". Here is a nice example: let $X=\mathbb{R}/\mathbb{Z}$, $f(x):=2x$, let $U_1=[0,\frac{1}{2})+\mathbb{Z}$ and $U_2=X\setminus U_1$. What can we say about the manner in which the trajectory $f^n(\sqrt{2})$ visits $U_1$ and $U_2$? For example, does the trajectory spend an equal amount of time in both sets? Put another way, is $\sqrt{2}$ normal to base $2$? Dynamical methods can tell us that Lebesgue a.e. number is normal to base $2$, but they struggle badly for specific numbers. The difficulty of this problem is tied quite closely to the existence of many invariant measures for this map: Lebesgue measure is invariant, but so is the Dirac measure at $0$; there are many invariant measures supported on closed $f$-invariant proper subsets corresponding to restricted digit sets (e.g. numbers where "11" is not allowed in the binary expansion) and also many fully supported invariant measures which are singular with respect to Lebesgue measure (and indeed with respect to one another). Each invariant measure contributes a set of points with its own particular, distinct dynamical behaviour, and given an arbitrary point we have no obvious way of knowing which of these different dynamical behaviours prevails.
There is only one situation in which this dynamical problem becomes easy: when a unique $f$-invariant Borel probability measure exists. In this case the convergence in the Birkhoff theorem is uniform[*] over all $x$, and the problem of distinguishing which invariant measure (if any) characterises the behaviour of the trajectory $f^n(x)$ vanishes completely. An example would be the leading digit of $2^n$: take $X=\mathbb{R}/\mathbb{Z}$, $U_k=[\log_{10}k,\log_{10}(k+1))+\mathbb{Z}$ for $k=1,\ldots,9$, $f(x):=x+\log_{10}2$, then the leading digit of $2^n$ is $k$ if and only if $f^n(0) \in U_k$. The irrationality of $\log_{10} 2$ may be applied to show that Lebesgue measure is the only $f$-invariant Borel probability measure, and using the uniform ergodic theorem we can simply read off the average time spent in each $U_k$ as the Lebesgue measure of that interval.
So how does this affect $f(x):=x(x+1)/2$ on $\mathbb{Z}_2$? Well, $x=1$ and $x=0$ are both fixed points and carry invariant Borel probability measures $\delta_1$ and $\delta_0$. So the dynamical system is not uniquely ergodic, and the problem of determining which of the (presumably many) invariant measures characterises the trajectory of the initial point $2$ has no obvious solution.
[*] to obtain uniform convergence in this case $\phi$ must be a continuous function. By upper/lower approximation we can deduce pointwise convergence to $\int \phi\,d\mu$ for all $x\in X$ in the case where $\phi$ is upper or lower semi-continuous, for example where $\phi$ is the characteristic function of a closed or open set.