I claim $(x,y,z)=(45,2,1)$ is the only solution. Note that if $y=1$ then $x^2\equiv 3\pmod{4}$; thus $y\ge 2$. Assume $y=2$. We have $$(x-2)(x+2) = 43^z \cdot 47^z.$$ Noticing $(x-2,x+2)=1$, it follows $x-2=43^z$ and $x+2=47^z$. That is, $47^z - 43^z=4$. From here, we immediately obtain $z=1$, yielding the triple above. Now assume $y\ge 3$. Using modulo $8$, it follows $z$ is even, set $z=2k$ with $k\ge 1$ to arrive at $$\bigl(x-2021^k\bigr)\bigl(x+2021^k\bigr) = 2^y.$$ In particular, (a) $x\pm 2021^k$ are both even; and (b) if $d\mid x\pm 2021^k$ then $d\mid 2\cdot 2021^k$. Consequently, $d=2$. That is, we must have $$2\cdot 2021^k + 2 = 2^{y-1} \implies 2021^k + 1 = 2^{y-2}.$$ Clearly, $y>4$. Inspecting modulo $4$, however, this is a clear contradiction.

grupyorum wrote: I claim $(x,y,z)=(45,2,1)$ is the only solution. Note that if $y=1$ then $x^2\equiv 3\pmod{4}$; thus $y\ge 2$. Assume $y=2$. We have $$(x-2)(x+2) = 43^z \cdot 47^z.$$ Noticing $(x-2,x+2)=1$, it follows $x-2=43^z$ and $x+2=47^z$. That is, $47^z - 43^z=4$. From here, we immediately obtain $z=1$, yielding the triple above. Now assume $y\ge 3$. Using modulo $8$, it follows $z$ is even, set $z=2k$ with $k\ge 1$ to arrive at $$\bigl(x-2021^k\bigr)\bigl(x+2021^k\bigr) = 2^y.$$ In particular, (a) $x\pm 2021^k$ are both even; and (b) if $d\mid x\pm 2021^k$ then $d\mid 2\cdot 2021^k$. Consequently, $d=2$. That is, we must have $$2\cdot 2021^k + 2 = 2^{y-1} \implies 2021^k + 1 = 2^{y-2}.$$ Clearly, $y>4$. Inspecting modulo $4$, however, this is a clear contradiction. why is z an even number?

Because if $z$ is odd number then $x^2\equiv 0+5^z\equiv 5^{2k+1}\equiv(25)^k\cdot 5\equiv 5\pmod{8}$ which is absurd.

What is d?

d is a random divisor of x+-2021^k and 2*2021^k