Given $$\,\left| {{z^2} - 1} \right| = {\left| z \right|^2} + 1$$,

By squaring both sides we get,

$${\left| {{z^2} - 1} \right|^2}$$ = $${\left( {{{\left| z \right|}^2} + 1} \right)^2}$$

$$ \Rightarrow $$ $$\left( {{z^2} - 1} \right)$$$$\overline {\left( {{z^2} - 1} \right)} $$ = $${\left( {{{\left| z \right|}^2} + 1} \right)^2}$$ [ as $${{{\left| z \right|}^2}}$$ = $$z\overline z $$ ]

$$ \Rightarrow $$ $$\left( {{z^2} - 1} \right)$$$$\left( {{{\left( {\overline z } \right)}^2} - 1} \right)$$ = $${\left( {{{\left| z \right|}^2} + 1} \right)^2}$$

$$ \Rightarrow $$ $${\left( {z\overline z } \right)^2}$$ $$-$$ $${{z^2}}$$ $$-$$$${{{\left( {\overline z } \right)}^2}}$$ $$+$$ 1 = $${\left| z \right|^4}$$ $$+$$ 2$${{{\left| z \right|}^2}}$$ $$+$$ 1

$$ \Rightarrow $$ $${\left| z \right|^4}$$ $$-$$ $${{z^2}}$$ $$-$$$${{{\left( {\overline z } \right)}^2}}$$ $$+$$ 1 = $${\left| z \right|^4}$$ $$+$$ 2$${{{\left| z \right|}^2}}$$ $$+$$ 1

$$ \Rightarrow $$ $${{z^2}}$$ $$+$$$${{{\left( {\overline z } \right)}^2}}$$ $$+$$ 2$${z\overline z }$$ = 0

$$ \Rightarrow $$ $${\left( {z + \overline z } \right)^2}$$ = 0

$$ \Rightarrow $$ $${z + \overline z }$$ = 0

$$ \Rightarrow $$ $$z$$ = $$-$$ $${\overline z }$$

If $$z$$ = x + iy

then $${\overline z }$$ = x - iy

$$\therefore$$ x + iy = - (x - iy)

$$ \Rightarrow $$ x + iy = - x + iy

$$ \Rightarrow $$ x = 0

$$\therefore$$ z is purely imaginary.

So, it is lie on the imaginary axis.