Find all positive integers $d$ for which there exist polynomials $P(x)$ and $Q(x)$ with real coefficients such that degree of $P$ equals $d$ and
$$P(x)^2+1=(x^2+1)Q(x)^2.$$
1. **Initial Setup and Degree Comparison:**
   We start with the given equation:
   \[
   P(x)^2 + 1 = (x^2 + 1)Q(x)^2
   \]
   Let the degree of \(P(x)\) be \(d\). Comparing the degrees on both sides, we see that:
   \[
   2d = 2(d-1) + 2
   \]
   Simplifying, we get:
   \[
   2d = 2d - 2 + 2 \implies \deg Q = d - 1
   \]

2. **Case 1: \(d\) is Even:**
   Suppose \(d\) is even. Then \(d-1\) is odd, so \(Q(x)\) has a real root \(x_0\). Substituting \(x_0\) into the original equation:
   \[
   P(x_0)^2 + 1 = (x_2 + 1)Q(x_0)^2
   \]
   Since \(Q(x_0) = 0\), we get:
   \[
   P(x_0)^2 + 1 = 0
   \]
   This is impossible because \(P(x_0)^2 + 1\) is always positive. Therefore, there are no solutions for even \(d\).

3. **Case 2: \(d\) is Odd:**
   We need to show that all odd \(d\) work. We start with the known solution:
   \[
   P(x) = x, \quad Q(x) = 1
   \]
   We aim to find polynomials \(A, B, C, D\) such that if \((P, Q)\) is a solution, then so is \((AP + BQ, CP + DQ)\). We want:
   \[
   (AP + BQ)^2 - (x^2 + 1)(CP + DQ)^2 = P^2 - (x^2 + 1)Q^2
   \]
   Expanding both sides, we get:
   \[
   \left[A^2 - (x^2 + 1)C^2\right]P^2 + 2\left[AB - (x^2 + 1)CD\right]PQ + \left[B^2 - (x^2 + 1)D^2\right]Q^2 = P^2 - (x^2 + 1)Q^2
   \]
   This is equivalent to:
   \[
   A^2 - (x^2 + 1)C^2 = 1, \quad AB = (x^2 + 1)CD, \quad B^2 - (x^2 + 1)D^2 = -(x^2 + 1)
   \]

4. **Finding Suitable Polynomials:**
   Trying \(C(x) = 2x\) and \(A(x) = 2x^2 + 1\), we check if they satisfy the equations:
   \[
   A^2 - (x^2 + 1)C^2 = (2x^2 + 1)^2 - (x^2 + 1)(2x)^2 = 4x^4 + 4x^2 + 1 - 4x^4 - 4x^2 = 1
   \]
   This works. Now, for \(B\) and \(D\), we try:
   \[
   B = 2x(x^2 + 1), \quad D = 2x^2 + 1
   \]
   Checking the second equation:
   \[
   AB = (2x^2 + 1)2x(x^2 + 1) = 2x(x^2 + 1)(2x^2 + 1) = (x^2 + 1)2x(2x^2 + 1) = (x^2 + 1)CD
   \]
   This works. Finally, checking the third equation:
   \[
   B^2 - (x^2 + 1)D^2 = [2x(x^2 + 1)]^2 - (x^2 + 1)(2x^2 + 1)^2 = 4x^2(x^2 + 1)^2 - (x^2 + 1)(4x^4 + 4x^2 + 1) = -(x^2 + 1)
   \]
   This also works.

5. **Inductive Step:**
   If \((P, Q)\) is a solution, then so is:
   \[
   ((2x^2 + 1)P + 2x(x^2 + 1)Q, 2xP + (2x^2 + 1)Q)
   \]
   Each time, the degrees of \(P\) and \(Q\) increase by 2, so all odd \(d\) are possible.

\(\blacksquare\)

The final answer is \( \boxed{ d } \) is odd.