Solutions with x,y positives integer for ${x}^{4}+4\,{x}^{3}+6\,{x}^{2}+4\,x+16={y}^{2}$
Rewrite as $$(x^2+2x+1)^2+15=y^2\implies y^2-A^2=15$$
for $A=x^2+2x+1\in \mathbb Z$
But then $(y-A)(y+A)=15$ so we seek divisors of $15$, and there are only a few cases to check.