1. **State the problem:** Solve for all values of $x$ in the equation $$\log_{x+3}(3x^2 + 1) = 2.$$\n\n2. **Recall the logarithm definition:** The equation $\log_a b = c$ means $$a^c = b,$$ where $a > 0$, $a \neq 1$, and $b > 0$.\n\n3. **Apply the definition:** Here, $a = x+3$, $b = 3x^2 + 1$, and $c = 2$. So, $$ (x+3)^2 = 3x^2 + 1. $$\n\n4. **Expand the left side:** $$ (x+3)^2 = x^2 + 6x + 9. $$\n\n5. **Set up the quadratic equation:** $$ x^2 + 6x + 9 = 3x^2 + 1. $$\n\n6. **Bring all terms to one side:** $$ x^2 + 6x + 9 - 3x^2 - 1 = 0 \implies -2x^2 + 6x + 8 = 0. $$\n\n7. **Simplify by dividing by -2:** $$ \cancel{-2}x^2 + \cancel{6}x + \cancel{8} = 0 \implies x^2 - 3x - 4 = 0. $$\n\n8. **Factor the quadratic:** $$ (x - 4)(x + 1) = 0. $$\n\n9. **Solve for $x$:** $$ x = 4 \quad \text{or} \quad x = -1. $$\n\n10. **Check domain restrictions:**\n- Base of logarithm $x+3 > 0 \implies x > -3$.\n- Base $x+3 \neq 1 \implies x \neq -2$.\n- Argument $3x^2 + 1 > 0$ always true for all real $x$.\n\nBoth $x=4$ and $x=-1$ satisfy $x > -3$ and $x \neq -2$.\n\n**Final answer:** $$\boxed{x = 4 \text{ or } x = -1}.$$