$4m^4-1,4m^4,4m^4+1$ are almost a squares for large $m$ ($4m^4+1=(2 m^2 - 2 m + 1) (2 m^2 + 2 m + 1)$ and $\dfrac{2 m^2 - 2 m + 1} {2 m^2 + 2 m + 1}>0.99 $ for $ m\ge 199$) Now, $4m^4+2$ must be almost a square. Let $4m^4+2=(2m^2-2a)(2m^2+b)$ $m^2=\dfrac{ab+1} {b-2a}$. Let $b-2a=1$ $m^2=2a^2+a+1$ This Pell's equation has infinitely many solutions starting with $(0,1);(1,2)$ and $4m^4+2=(4 a^2 + 2) (4 a^2 + 4 a + 3)$ Тherefore there are infinitely many consecutive quadruples of almost squares.