How Many Squares?

We can write $n^4 + 6n^3 + 13n^2 + 12n + 7 = (n^2 + 3n + 2)^2 + 3.$ The only squares that differ by 3 are 1 and 4, and these are easy to rule out. Therefore, there are no such positive integers $n$ .

A perfect square gives a remainder of $0$ or $1$ when divided by $4$ .

By Quotient Remainder Theorem

If $n \equiv 0 \pmod {4}$ , $n^4 + 6n^3 + 13n^2 + 12n + 7 \equiv 3 \pmod{4}$ , which is not allowed

If $n \equiv 1 \pmod {4}$ , $n^4 + 6n^3 + 13n^2 + 12n + 7 \equiv 3 \pmod{4}$ , which is not allowed

If $n \equiv 2 \pmod {4}$ , $n^4 + 6n^3 + 13n^2 + 12n + 7 \equiv 3 \pmod{4}$ , which is not allowed

If $n \equiv 3 \pmod {4}$ , $n^4 + 6n^3 + 13n^2 + 12n + 7 \equiv 3 \pmod{4}$ , which is not allowed

Since we have exhausted all cases, there is no $n$ such that $n^4 + 6n^3 + 13n^2 + 12n + 7$ is a perfect square.