The equation provided is:

$3^{x+2} - 5^x = 3^{x+4} - 5^{x+2}.$

Let us solve step by step:

Step 1: Simplify the exponents

We know the following properties of exponents: $a^{x+y} = a^x \cdot a^y.$ Apply this to rewrite terms:

$3^{x+2} = 3^x \cdot 3^2, \quad 3^{x+4} = 3^x \cdot 3^4, \quad 5^{x+2} = 5^x \cdot 5^2.$

Substitute back into the equation: $3^x \cdot 3^2 - 5^x = 3^x \cdot 3^4 - 5^x \cdot 5^2.$

Step 2: Factor out $3^x$ and $5^x$

First, simplify constants: $3^2 = 9, \quad 3^4 = 81, \quad 5^2 = 25.$

The equation becomes: $9 \cdot 3^x - 5^x = 81 \cdot 3^x - 25 \cdot 5^x.$

Rearrange terms: $9 \cdot 3^x - 81 \cdot 3^x = 25 \cdot 5^x - 5^x.$

Factor each side: $3^x (9 - 81) = 5^x (25 - 1).$

Simplify: $3^x (-72) = 5^x (24).$

Step 3: Divide through by $24$ (or simplify further)

$\frac{3^x}{5^x} = -\frac{24}{72}.$

We observe that this simplifies further for real solutions where ratios can balance logarithmically. To discuss with cleaner logarithms ...Let us simplify and continue from the last result:

The simplified equation was:
$3^x \cdot (-72) = 5^x \cdot 24.$

This can be rewritten as:
$\frac{3^x}{5^x} = -\frac{24}{72} = -\frac{1}{3}.$

Taking $\log$ on both sides doesn't lead to valid real solutions because negative arguments of logs are undefined.

Observations

This equation does not have real solutions because no exponential ratio involving $3^x$ and $5^x$ will equate to a negative value (like $-1/3$).

Thus, the equation has no real solutions.