2 backticks

Agreed, nice spirals. This is not meant as a complete answer. Heuristically and experimentally, it is very likely that there are infinitely many $x,y\in\mathbb{Z}$ such that $x^2+y^2$, $(x+2)^2+y^2$, $x^2+(y+2)^2$ and $(x+2)^2+(y+2)^2$ are simultaneously prime. (For example, PARI gave $136$ such $(x,y)$ with $1\leq x,y\leq 1000$ and $1330$ such $(x,y)$ with $1\leq x,y \leq 5000$.) Such a result will give Gaussian primes at the corners of a square of side length two and hence infinitely many of the simplest type of cycle. Though it may not directly solve your question, it should be noted that you do get infinitely many squares with Gaussian primes at the corners if you allow the lengths of the sides of the squares to vary, due to the following result:

T. Tao, The Gaussian primes contain arbitrarily shaped constellations, J. d.Analyse Mathematique 99 (2006), 109--176.

Also, heuristically, if you believe the kind of heuristics behind the Hardy-Littlewood $k$-tuple conjecture, the same kind of reasoning should give you the conjectural asymptotics for

$$\sum_{x,y\in sum_{\substack{x,y\in \mathbb{Z}}\Lambda(x^2+y^2)\Lambda((x+2)^2+y^2)\Lambda(x^2+(y+2)^2)\Lambda((x+2)^2+(y+2)^2). mathbb{Z}\\1 \leq x,y\leq N}}\Lambda(x^2+y^2)\Lambda((x+2)^2+y^2)\Lambda(x^2+(y+2)^2)\Lambda((x+2)^2+(y+2)^2).$$

1

Agreed, nice spirals. This is not meant as a complete answer. Heuristically and experimentally, it is very likely that there are infinitely many $x,y\in\mathbb{Z}$ such that $x^2+y^2$, $(x+2)^2+y^2$, $x^2+(y+2)^2$ and $(x+2)^2+(y+2)^2$ are simultaneously prime. (For example, PARI gave $136$ such $(x,y)$ with $1\leq x,y\leq 1000$ and $1330$ such $(x,y)$ with $1\leq x,y \leq 5000$.) Such a result will give Gaussian primes at the corners of a square of length two and hence infinitely many of the simplest type of cycle. Though it may not directly solve your question, it should be noted that you do get infinitely many squares with Gaussian primes at the corners if you allow the lengths of the sides of the squares to vary, due to the following result:

T. Tao, The Gaussian primes contain arbitrarily shaped constellations, J. d.Analyse Mathematique 99 (2006), 109--176.

Also, heuristically, if you believe the kind of heuristics behind the Hardy-Littlewood $k$-tuple conjecture, the same kind of reasoning should give you the conjectural asymptotics for

$$\sum_{x,y\in \mathbb{Z}}\Lambda(x^2+y^2)\Lambda((x+2)^2+y^2)\Lambda(x^2+(y+2)^2)\Lambda((x+2)^2+(y+2)^2).$$